# Solution Set 1


The solutions manual for the exercises are primarily given in four parts: Question, Answer, Hint, and Solution. Not all problems have all four of these components.

## 1.1

Calculate the result of each of the following and record answer with the proper number of significant figures.

1. 13.4 + 9.8 + 55.99
2. 340.5 * 419.23
3. 19.45 / 14.23
4. 7 - 2.8

1. 79.2
2. 1.427 x 105
3. 1.367
4. 4

Hint: Significant Digits

Solution: When adding and subtracting numbers, the answer may not contain more decimal places than the least accurate number. When multiplying or dividing numbers, record the answer to the number of significant figures as the least accurate number.

## 1.2

Calculate and express the result in the correct number of significant figures.

1. (1.602 x 10-19 ) x (2.15 x 103)
2. (3.1667 x 1023 ) x (9.018 x 10-9) / (7.52 x 104)
3. 1.714 x 10-1 + 1.96 x 10-3 + 1.255 x 10-2
4. (7.00743 - 7.00521) / (5.9831 x 1019 )
5. (4.832 x 10-5 - 4.760 x 10-5) / (4.832 x 10-5) x 100 (100 is exact)
6. (9.02 x 103 + 8.99 x 103 + 1.065 x 104) / 3 (3 is exact)

1. 3.44 x 10-16
2. 3.80 x 1010
3. 1.859 x 10-1
4. 3.71 x 10-23
5. 1.5
6. 9.55 x 103

Solution:

1. 3.4443 x 10-16 = 3.44 x 10-16 (3 sig. figs)
2. 3.797513378 x 1010 = 3.80 x 1010 (3 sig. figs)
3. 1.714 x 10-1 = 0.1714 (4 dec. places)

1.96 x 10-3 = 0.00196 (5 dec. places)

1.255 x 10-2 = 0.01255 (5 dec. places)

the answer should have 4 decimal places (the least amount of decimal places)

1.714 x 10-1 + 1.96 x 10-3 + 1.255 x 10-2 = 0.18591 = 0.1859 = 1.859 x 10-1

1. (7.00743 - 7.00521) / (5.9831 x 1019 ) = 0.00222 / (5.9831 x 1019) = 3.7104511 x 10-23 = 3.71 x 10-23 (3 sig. figs)
2. (4.832 x 10-5 - 4.760 x 10-5) / (4.832 x 10-5) x 100 = (7.2 x 10-7) / (4.832 x 10-5) x 100 = 1.4906623 = 1.5 (2 sig.figs)
3. (9.02 x 103 + 8.99 x 103 + 1.065 x 104) / 3 = (2.8660 x 104 ) / 3 = 9.553333333 x 103 = 9.55 x 103 (3 sig. figs; do not round up until the end of calculations)

## 1.3

Convert the following temperatures to Kelvin and Fahrenheit degrees.

1. room temperature: 20.0 oC
2. the boiling point of water: 100.oC
3. the temperature of human: 37.0 oC

1. 293 K; 68.0 oF
2. 373 K; 212oF
3. 310.K; 98.6oF

Solution:

1. 20.0+273=293K; 20.0*9/5+32=68.0oF
2. 100+273=373K; 100*9/5+32=212oF
3. 37.0+273=310K; 37.0*9/5+32=98.6oF

Hint: Temperature

## 1.4

Using the following table, calculate the volume of 50.0 g of each of the following substances at 20.0o C. What does the volume of oxygen gas tell you about the properties of gasses?

 Substance Physical State Density (g/cm3) Ethanol Liquid 0.789 Iron Solid 7.87 Oxygen Gas 0.00133
1. Oxygen gas
2. Iron
3. Ethanol

1. 3.76x104 cm3
2. 6.35 cm3
3. 63.4 cm3

The volume of 50.0 g of O2 (g) is much larger than the other volumes, indicating that gasses consist of mostly empty space.

Solution: Converting from mass to volume requires that you use an inverse density value as well as dimensional analysis.

$(50.0\; g\; Ethanol) \times \dfrac{1\;cm^3}{0.789\;g} = 63.4\; cm^3.$

$(50.0\; g\; Iron) \times \dfrac{1\;cm^3}{7.87\;g} = 6.35\; cm^3.$

$(50.0\; g\; Oxygen) \times \dfrac{1\;cm^3}{0.00133\;g} = 3.76\; \times 10^4 cm^3.$

Gasses tend to have much larger volumes than solids and liquids because they fill the container that they are placed in. Gasses exhibit weak intermolecular forces, causing them to expand readily and occupy a large volume.

## 1.5

A student gets three thermometers from the lab. He notices that these thermometers use different units for measurements. They are oF, oC, and K. If he puts these thermometers into three water baths and observe a 10 degree change for each thermometer, which water bath has the smallest change in temperature ?

Answer: The water bath using oF thermometer

Hint: Consider the difference between two known temperature values whose difference is 10 degrees.

Solution: For convenience, choose to convert zero and 10 degrees for each unit to degrees C, then calculate the difference (final T - initial T).

Fahrenheit: For example, 0oF is about -18oC, which can be solving using the formula oF = 9/5oC + 32, solving for oC to get oC=(5/9)(oF-32) , and then plugging in 0 for oF. 10oF is about -12oC, using the same method. Thus, the change in temperature in oC, is -12oC-(-18​oC) = 5.6oC for a 10 degree shift in the Fahrenheit scale.

Kelvin: To convert Kelvin to oC, the formula K = 273.15 + oC is used, and for 0K, it results in -273oC; for 10K, it results in -263oC. Thus, the change in temperature in oC, is -263oC-(-273oC) = 10oC for a 10 degree shift in the Kelvin scale.

Celsius: Finally, the difference in temperature between 0oC and 10oC is 10oC in the Celsius scale.

Result: Comparing the 10 degree differences in the three units, 5.6oC is the smallest, which originated from a 10 degree difference in the Fahrenheit scale.

## 1.6

What is the main difference between each of the following pairs?

1. An elemental substance (element) and a compound
2. Physical and chemical changes
3. Homogeneous and heterogeneous mixtures

1. An element (or elemental substance) is composed entirely of a single element. A compound has atoms bonded that are different elements.
2. A physical change is a change in the state of the matter without changing the substance itself, like ice changing into water or a foil pulled into a wire or a metal ball hammered into a sheet. A chemical change is a change where the original matter changes into different matter. An example of these is making and baking cookies. Mixing flour and sugar is a physical change because the individual grains are still made of flour and sugar. Baking cookies is a chemical change because the heat alters the bonds between atoms to create something new .
3. A homogeneous mixture is one in which, given any part of the mixture, it has the same concentration (amount of matter per volume) as any other part of the whole, like a brewed cup of coffee or a round of cheese. This is not the case for a heterogeneous mixture, which can be represented by a bowl of mixed jelly beans or gravel.

Hint:

1. One contains same elements, the other contains different elements.
2. Melting compared to burning.
3. Potting soil compared to tea.

## 1.7

Define whether each of the following is a pure substance or a mixture.

2. copper
3. bronze
4. silk
5. ice
6. sand
7. magnesium
8. solid potassium iodide (KI)
9. honey

a) mixture

b) pure

c) mixture

d) mixture

e) pure

f) mixture

g) pure

h) pure

i) mixture

## 1.8

Are these chemical or physical changes?

1. A nail rusting
2. Toast burning
3. Water boiling
4. Ice cream melting

1. Chemical change
2. Chemical change
3. Physical change
4. Physical change

Solution:

1. A nail rusting is a chemical change because a new substance,rust, has formed.
2. Burnt toast is a chemical change because a new substance, burned toast, was created.
3. Water boiling is a physical change because a new substance has not formed. The molecules of water are still composed of two hydrogen atoms and one oxygen atoms. Simply removing the heat source will cause the water to condense.
4. Melting of ice cream is a physical change because no new substance formed when the ice cream melted. Instead, it changed state from a solid to a liquid.

## 1.9

The actual amount of magnesium in a sample is 21.14%. A student analyzed the percent composition of magnesium in the sample and obtained the results: 20.36%, 20.37%, 20.34%, 20.37%. Explain the accuracy and precision of these results.

Answer: The data the student collected about the amount of magnesium in the sample is precise, but not accurate.

Solution: The data that the student collected is precise because the four percentages are close to one another. The data is not accurate because the numbers collected are significantly lower than the actual percentage of magnesium in the sample.

## 2.1

Both J.J Thomson and Ernest Rutherford had different understandings of how the structure of an atom actually looked like. What did each of them postulate about the structure of the atom?

J.J Thomson postulated that an atom was a diffused cloud with tiny particles of positive charge and electrons scattered randomly within that cloud. Thomson called this the plum pudding model.

Ernest Rutherford postulated that an atom must contain a dense center of positive charge with negatively charged electrons surrounding it at a large distance relative to the size of the area the positive charge is contained in (now know as the nucleus). This structure of the atom was called the nuclear atom.

Hint: Read about the experiments that J.J Thomson and Ernest Rutherford did.

## 2.2

Give the symbol of each ion for the following :

1. An ion with 20 protons, 20 neutrons, 18 electrons
2. An ion with 28 protons, 30 neutrons, 26 electrons
3. An ion with 28 protons, 30 neutrons, 24 electrons

1. $$\ce{^{40}Ca^{2+}}$$
2. $$\ce{^{58}Ni^{2+}}$$
3. $$\ce{^{58}Ni^{4+}}$$

Hint: What subatomic particles make up the atomic mass of an ion? http://chemwiki.ucdavis.edu/Physical_Chemistry/Atomic_Theory/The_Atom

Solution:

1. An element with 20 protons signifies Calcium. The atomic mass is the sum of the protons and neutrons, which is 20 protons + 20 neutrons. Therefore the atomic mass of the ion is 40. To find the charge, subtract the number of electrons from the number of protons, 20 – 18, therefore the charge of the ion is +2.
2. An element with 28 protons signifies Nickel. Adding the number of protons and neutrons together gives the atomic mass of this ion to be 58. Subtracting the electrons from the protons gives the charge of the ion to be 2+ because 28 – 26 = +2.
3. An element with 28 protons signifies Nickel. Adding the number of protons and neutrons together gives the atomic mass of this ion to be 58. Subtracting the electrons from the protons gives the charge of the ion to be 4+ because 28 – 24 = +4.

## 2.3

Bromine has two stable isotopes: $$\ce{^{79}Br}$$ and $$\ce{^{81}Br}$$. The mass of 81Br is 80.92 u. The proportion of 79Br is 50.69%. The average atomic mass is 79.91 u. What is the mass of $$\ce{^{79}Br}$$?

Answer mass of $$\ce{^{79}Br}$$ is 78.92 u.

Hint: Consider the definiton of atomic mass.

http://chemwiki.ucdavis.edu/Wikitext...s#Average_Mass

Solution:

80.92u*(1-0.5069) + X*0.5069 =79.91u

X=78.92u

## 2.4

Choose the true statement(s) and correct the wrong statement(s).

1. Particles in a nucleus of an atom are positive charged.
2. In an atom, electrons orbit around the nucleus.
3. The mass of the nucleus is approximately equals to the add-up mass of protons and neutrons.
4. The number of electrons equals to the number of neutrons.
5. Mass of an electron approximately equals to the mass of a proton.

Answer Only b and c are correct.

• Neutrons are neutral
• The number of electrons equals to the number of protons.
• Mass of an electron is much lighter than a proton.

Solution: Atom is a neutral, which is made of electrons, protons and neurons. Electron has negative charge, proton has positive charge, and neuron is neutral. In an atom, number of electron is equal to number of protons so that the atom is neutral because negative charge and positive charge cancel out each other. While the number of neurons may be different for an element so that isotopes formed. Since mass of electron is relatively extremely small, mass of an atom usually equal to the add-up mass of protons and neurons. Electron is much lighter than proton and neutron in an atom. The electrons orbit around the nucleus

## 3.1

Give systematic names for the compounds or write the formula for the given names of the compounds.

1. $$HgO$$
2. $$FeI_3$$
3. $$SiCl_4$$
4. $$NO_2$$
5. tin(II) nitride
6. copper(II) oxide
7. mercury (I) oxide
8. sodium dichromate

Answer a. mercury (II) oxide b. iron (III) iodide c. silicon chloride. e. $$Sn_3N_2$$ f. $$CuO$$ g. $$Hg_2O$$ h, $$Na_2Cr_2O_7$$
Hint: http://chemwiki.ucdavis.edu/Physical_Chemistry/Quantum_Mechanics/Atomic_Theory/Chemical_Compounds/Nomenclature_of_Inorganic_Compounds

Solution:
nonmetal covalent compound, the rule to name it is: prefix+ nonmetal+ prefix+ nonmetal.
an ionic compound with variable charge metal, the rule to name it is: metal name (number of charge of metal) + nonmetal ending with –ide.
an ionic compound with fixed charge metal, so omit the number of charge of metal.

## 3.2

ASSIGN the CORRECT IONIC formula for each of the following compounds:

1. sodium chloride
2. tin(II) fluoride
3. magnesium nitride
4. aluminum oxide
5. mercury(I) sulfide
6. silver bromide

Hint: Can you tell by looking their names? Do they all have the same charges?

1. $$NaCl$$
2. $$SnF_2$$
3. $$Mg_3N_2$$
4. $$Al_2O_3$$
5. $$Hg_2S$$
6. $$AgBr$$

Solution: To form a neutral compound from anion and cation, the positively and negatively charges must be balanced to zero. Ex:

$Na^+ + Cl^- \rightarrow NaCl_{(s)}$

$2 Al^{3+} + 3 O^{2-} \rightarrow Al_2O_{3\; (s)}$

## 3.3

State the name of the following compounds.

1. $$HgO$$
2. $$HNO_2$$
3. $$HClO_3$$
4. $$KCl$$
5. $$OF_2$$
6. $$LiMnO_4$$
7. $$CaS_2O_3$$
8. $$Fe_2O_3$$
9. $$CO$$
10. $$SO_4$$
11. $$NaOH$$
12. $$HCN$$

1. Mercury(II) oxide
2. Nitrous acid
3. Chloric acid
4. Potassium chloride
5. Oxygen difluoride
6. Lithium permanganate
7. Calcium thiosulfate
8. Iron(III) oxide
9. Carbon monoxide
10. Sulfur tetraoxide
11. Sodium hydroxide
12. Hydrocyanic acid

## 3.4

Give names of the following ions or acids.

1. $$CIO^-$$
2. $$ClO_2^-$$
3. $$ClO_3^-$$
4. $$ClO_4^-$$
5. $$HClO$$
6. $$HClO_2$$
7. $$HClO_3$$
8. $$HClO_4$$

1. $$CIO^-$$ hypochlorite ion
2. $$ClO_2^-$$ chlorite ion
3. $$ClO_3^-$$ chlorate ion
4. $$ClO_4^-$$ perchlorate ion
5. $$HClO$$ hypochlorous acid
6. $$HClO_2$$ chlorous acid
7. $$HClO_3$$ chloric acid
8. $$HClO_4$$ perchloric acid

Solution:
The name of acid or chlorine oxyanions vary as the oxidation sate of I change.

For the chlorine oxyanions:

• $$Cl^+$$: prefix with hypo-
• $$Cl^{+3}$$: iodite ion
• $$Cl^{+5}$$: suffix with –ate
• $$Cl^{+7}$$: prefix with per- and suffix with –ate

For the acid:
With respect to the oxidation number of Cl, the rule of the naming respectively:

• Prefix with hypo- and suffix with –ous acid.
• chlorous acid
• Suffix with –dic acid
• Prefix with per- and suffix with –dic acid.

## 3.5

There are some chemical stuffs that are unlabeled in a chemistry lab. Please write down the formula for each of the followings for the students:

1. nickel(II) iodide
2. aluminum perchlorate
3. ammonia
4. magnesium dichromate
5. ammonium phosphate
6. xenon tatrafluoride

Hint: Can you tell by looking their names? Do they all have the same charges? When do you use Greek prefix?

1. $$NiI_2$$
2. $$Al(CIO_4)_3$$
3. $$NH_3$$
4. $$MgCr_2O_7$$
5. $$(NH_4)_3PO_4$$

*$$NH_4^+$$ here is an individual molecule. We use parentheses to indicate the difference between a molecule and its quantity.

1. $$XeF_4$$

## 3.6

Write the molecular formula for all of the following COMPOUNDS:

1. Barium Chlorate
2. Carbon monoxide
3. Carbon dioxide
4. Carbon Tetrachloride
5. Germanium (IV) Sulfide
6. Germanium (II) Sulfide
7. Lithium Bromide

1. $$BaClO_3$$
2. $$CO$$
3. $$CO_2$$
4. $$CCl_4$$
5. $$GeS_2$$
6. $$GeS$$
7. $$LiBr$$

Hint: Remember to balance the charges given off by each individual ion.

Solution:

1. The Chlorate ion ($$ClO_3$$) has a -2 charge and the barium ion has a +2 charge.
2. Carbon monoxide has an overall 2- charge, unlike carbon dioxide which is neutral
3. Carbon has a +4 charge whereas Oxygen has a -2 charge.
4. Carbon has a +4 charge whereas Chlorine has -1 charge because it is a halogen.
5. Germanium (IV) has a +4 charge, thus two Sulfur ions are required to balance it.
6. Germanium (II) has a +2 charge, thus only one sulfur ion is needed to balance the charge of the Germanium.
7. Lithium has a +1 charge and bromide has a -1 charge

## 3.7

Give the chemical formula for each compound name:

1. lithium nitrite
2. antimony (III) sulfide
3. cobalt (III) hydroxide
4. tin (IV) iodide
5. arsenic acid
6. cesium telluride
7. perchloric acid
8. acetic acid
9. nickel (II) nitride
10. dinitrogen pentoxide
11. hydrofluoric acid
12. iron (III) oxide
1. LiNO2
2. Sb2S3
3. Co(OH)3
4. SnI4
5. H3AsO4
6. CsTe
7. HClO4
8. HC2H3O2
9. Ni3N2
10. N2O5
11. HF
12. Fe2O3

Hint: Link to hint for this question

Solution: Identify the oxidation states of each cation and anion, then form compounds with neutral charges. For example, $$Li^+$$ forms a 1:1 ratio with $$NO_2^-$$. When a compound is formed from covalent bonds you will see Greek numerical prefixes like di, tri, tetra, etc. as seen in dinitrogen pentoxide. [Answers shown above]

## 3.8

Why is $$Na_2 CrO_4$$ called Sodium Chromate and why is $$SnCrO_4$$ called Tin(II) Chromate ?

Hint: Reread Naming Type I Binary Compounds and Naming Type II Binary Compounds. http://chemwiki.ucdavis.edu/Under_Construction/SARIS/Liini/Towson/Naming_Inorganic_Compounds

Solution: The reason we call $$Na_2 CrO_4$$ Sodium Chromate is because in$$Na_2 CrO_4$$ there is only one type of sodium which is $$Na^+$$. On the other hand $$SnCrO_4$$ is called Tin(II) Chromate because $$Sn^{2+}$$ is a transition metal. That means there are different forms of Tin, which may be $$Sn^{2+}$$ or $$Sn^{4+}$$. As a consequence the charge of the cation is written in roman numerals to indicate which form of Tin we are talking about.

## 3.9

Give the systematic names of the substances indicated by the molecular formulas below :

1. $$CaCO_3$$
2. $$NH_4Cl$$
3. $$BaSO_3$$
4. $$Li_3PO_4$$
5. $$NaOH$$

1. Calcium Carbonate
2. Ammonium Chloride
3. Barium Sulfite
4. Lithium Phosphate
5. Sodium Hydroxide

Hint: Use an ion sheet to find the names of individual ions.

1. $$CO_3^{2-}$$ refers to the common ion Carbonate
2. $$NH_4^+$$ refers to the common ion Ammonia
3. $$SO_3^{2-}$$ is the common ion Sulfite
4. $$PO_4^{3-}$$ refers to the common ion Phosphate
5. $$OH^-$$ refers to the common ion of hydroxide.

## 3.10

Give the molecular formula of the compounds indicated below:

1. Magnesium Hydroxide
2. Sodium Nitrate
3. Acetic Acid
4. Silicon Tetrachloride
5. Magnesium Sulfate

1. $$Mg(OH)_2$$
2. $$NaNO_3$$
3. $$C_2H_4O_2$$
4. $$SiCl_4$$
5. $$MgSO_4$$

Hint: Remember to balance the charges of each individual ion.

Solution:

1. The Hydroxide ion has a charge of -1, and the magnesium atom has a charge of +2; thus two hydroxide ions balance out one Magnesium ion.
2. The Nitrate ion has a charge of -1, and the sodium ion has a charge of +1
3. This is a formula of a well-known acid found in vinegar and should be memorized
4. The name of the compound indicates four chloride ions through the pronoun –tetra
5. A Magnesium ion has a +2 charge, and a Sulfate ion has a -2 charge.

## 3.11

What are the empirical formula mass and molar mass of $$H_2O_2$$? What is the relationship between them?

Answer Empirical formula mass 34.02 g/mol, Molar mass 17.01 g/mol. The molar mass is based on the actual molecular formula while the empirical formula mass based on the empirical formula.

Hint: What is the definitions of empirical formula mass and molar mass?

Solution: Molar mass:

$16.00 \times 2 + 1.01 \times 2=34.02\; \text{g/mol}$

Empirical formula mass:

$1.01 + 12.01 = 17.01\; \text{g/mol}$

The empirical mass is the sum of atomic masses of the lowest ratio of all of the elements. The molar mass is the sum of the atomic masses of all of the elements. The molar mass is based on the actual molecular formula while the empirical formula mass based on the empirical formula. An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers. These two masses are always being related by a factor of a whole number.

## 3.12

 Compounds $$CH_5N$$ $$C_2H_6O_2$$ $$H_2SO_4$$ $$CH_3COCl$$ Mass(g) 157.5 # of Mole 1.404 # of Molecules $$4.022 \times 10^{23}$$ Total # of Atoms $$1.356 \times 10^{25}$$

 Compounds $$CH_5N$$ $$C_2H_6O_2$$ $$H_2SO_4$$ $$CH_3COCl$$ Mass(g) 43.61 157.5 65.52 252.5 # of Mole 1.404 2.538 0.6680 3.217 # of Molecules 8.455*1023 1.528*1024 $$4.022 \times 10^{23}$$ 1.937*1024 Total # of Atoms 5.918*1024 1.528*1025 2.816*1024 $$1.356 \times 10^{25}$$

Hint: The Mole and Avogadro's Constant

Solution:

a.MM: 31.06 g/mol

1.404*31.06=43.61g, 1.404*6.022*1023=8.455*1023, 8.455*1023*7=5.918*1024

b.MM: 62.07g/mol

157.5/62.07=2.538mol, 2.538* 6.022*1023=1.528*1024, 1.528*1024*10=1.528*1025

c.MM: 98.08g/mol

4.022*1023/(6.022*1023)=0.668mol, 0.668*98.08=65.52g, 4.022*1023*7=2.816*1024

d.MM: 78.49g/mol

1.356*1025/7=1.937*1024, 1.937*1024/(6.022*1023)=3.217mol, 3.217*78.49=252.5g

## 3.13

Ethylene glycol is a raw material in the manufacture of polyester fibers. It only contains hydrogen, carbon and oxygen. Their percentages are 9.763% H, 38.70% C and 51.55% O. If the molar mass of ethylene glycol is 62.07 g/mol, what is the molecular formula of ethylene glycol? If the molecular formula is different from its empirical formula, please write its empirical formula.

Answer Molecular formula $$C_2H_6O_2$$, Empirical formula $$CH_3O$$

Hint: What is the difference between molecular formula and empirical formula?

http://chemwiki.ucdavis.edu/Wikitext...cular_Formulas

Solution:

• H: 62.07(g/mol)*0.09763=6.060(g/mol) H, 6.060(g/mol)/1.01(g/mol)=6H.
• C: 62.07(g/mol)*0.3870=24.02(g/mol) C, 24.02(g/mol)/12.01(g/mol)=2C.
• O: 62.07(g/mol)*0.5155=32.00(g/mol) O, 32.00(g/mol)/16.00(g/mol)=2O.

## 3.14

A 5.00 g sample of an unknown compound with 82.14 g/mol is completely burned to give 16.08 g of carbon dioxide and 5.48 g of water. What is the empirical formula of this compound?

Answer$$C_6H_{10}$$

Hint: Determine the mass of C and H in the yield product, and then calculate the mass percent of C and H in the compound. After that, calculate the number of C and H atoms in the compound.

Solution: Since the product is carbon dioxide and water, we know that the compound is compose of two elements: carbon and hydrogen. However, we still don’t know whether it has oxygen or not because there exists a case that all of the oxygen from $$CO_2$$ is from the air.

Calculate the molar mass of $$CO_2$$, Molar mass of CO2=12.01+2x16.00=44.01g/mol

The fraction of C present by mass = (12.01 g C)/(44.01g CO2)

The mass of C in 16.08g of CO2 = (16.08g CO2) (12.01 g C)/(44.01g CO2)=4.388g CO2

This carbon is originally came from a 5.00g unknown compound, so the mass percent of C in this compound is (4.388g C)/(5.00 g compound)x100%=87.8%

Similarly, we can apply this calculating procedure for hydrogen.

Calculate the molar mass of H2O, Molar mass of H2O=1.008x2+16=18.016g/mol

The fraction of H present by mass = (2x1.008g H)/(18.016g H2O)

The mass of H in 5.48g of H2O = (5.48g H2O)(2x1.008g H)/(18.016g H2O) =0.613g H

The mass percent of H in this compound is (0.613g H)/ (5.00 g compound)x100%=12.2%

Since 87.7%+12.2% = 100%, we can conclude that there is no oxygen element in the unknown compound.

Calculate the number of carbon and hydrogen atoms in the 5.00g sample:

For every 100 g compound, there is 87.8g C and 12.2g H.

(87.8g C)(1 molC/12.01g C)=7.31molC

(12.2g H)(1 molH/1.008gH)=12.1molH

Find the smallest whole number ratio of atoms in this compound:

C: 7.31/7.31=1.00 H: 12.1/7.31=1.655 multiply both values by 3, we can get C3H5.

Empirical formula mass = 12.01x3+1.008x5=41.07g/mol

Molecular formula = (C3H5)(82.14g/mol/41.07g/mol)=C6H10

## 3.15

Glycerol is a sugar alcohol compound composed of only carbon, hydrogen, and oxygen. It is commonly used in pharmaceutical and food industry. Combustion of 1.79 grams of glycerol yields 2.57 g CO2 and 1.40 g $$H_2O$$. Find the empirical formula of glycerol.

Answer $$C_3H_8O_3$$.

Solution:

First, we find the percent mass of C, H, and O.

(2.57 g CO2) * (12.011 g C / 44.010 g CO2) = 0.701 g C

% C = (0.701 g C) / (1.79 g glycerol) * 100 = 39.2%

(1.40 g H2O) * [(2*1.00794 g H) / 18.0153 g H2O] = 0.157 g H

% H = (0.157 g H) / (1.79 g glycerol) * 100 = 8.77%

% O = 100% – 39.2% – 8.77% = 52.0%

Next, suppose we have 100. g of glycerol. Therefore, we have 39.2 g C, 8.77 g H, and 52.0 g O.

(39.2 g C) * (1 mol C / 12.011 g C) = 3.26 mol C

(8.77 g H) * (1 mol H / 1.00794 g H) = 8.70 mol H

(52.0 g O) * (1 mol O / 15.9994 g O) = 3.25 mol O

Then, we divide by the smallest mol. In this case, it is 3.25 mol O.

3.26 mol C / 3.25 mol O ≈ 1 mol C : 1 mol O

8.70 mol H / 3.25 mol O ≈ 2.68 mol H : 1 mol O

multiply all by 3, C: H: O = 3:8:3

Therefore, the empirical mass of glycerol is $$C_3H_8O_3$$.

## 3.16

Find the mass percent composition of the following molecules.

1. $$C_3H_8$$
2. $$CHCl_3$$
3. $$(C_2H_5)_2O$$

1. % C = 81.714%; % H = 18.2865
2. % C = 10.061%; % H = 0.844327%; % Cl = 89.095%
3. % C = 64.817%; %H = 13.598%; % O = 21.585%

Solution:

1. MM C3H8 = (3*12.011) + (8*1.00794) = 44.09652 g/mol

% C = (3*12.011 g/mol) / (44.09652 g/mol) * 100 = 81.714%

% H = 100% – 81.714% = 18.286%

1. MM CHCl3 = (1*12.011) + (1*1.00794) + (3*35.453) = 119.37794 g/mol

% C = (12.011 g/mol) / (119.37794 g/mol) * 100 = 10.061%

% H = (1.00794 g/mol) / (119.37794 g/mol) * 100 = 0.844327%

% Cl = 100% – 10.061% – 0.844327% = 89.095%

1. MM (C2H5)2O = (4*12.011) + (10*1.00794) + (1*15.9994) = 74.1228 g/mol

% C = (4*12.011 g/mol) / (74.1228 g/mol) * 100 = 64.817%

% H = (10*1.00794 g/mol) / (74.1228 g/mol) * 100 = 13.598%

% O = 100% – 64.817% – 13.598% = 21.585%

## 3.17

Ethylene glycol is a raw material in the manufacture of polyester fibers. It only contains hydrogen, carbon and oxygen. Their percentages are 9.763% H, 38.70% C and 51.55% O. If the molar mass of ethylene glycol is 62.07 g/mol, what is the molecular formula of ethylene glycol? If the molecular formula is different from its empirical formula, please write its empirical formula.

Answer: Molecular formula $$C_2H_6O_2$$, Empirical formula $$CH_3O$$

Hint: What is the difference between molecular formula and empirical formula?

http://chemwiki.ucdavis.edu/Wikitext...cular_Formulas

Solution:

• H: 62.07(g/mol)*0.09763=6.060(g/mol) H, 6.060(g/mol)/1.01(g/mol)=6H.
• C: 62.07(g/mol)*0.3870=24.02(g/mol) C, 24.02(g/mol)/12.01(g/mol)=2C.
• O: 62.07(g/mol)*0.5155=32.00(g/mol) O, 32.00(g/mol)/16.00(g/mol)=2O.

## 3.18

Indicate the oxidation state of the element assigned in the compounds below:

1. $$C$$ in $$CO_2$$
2. $$S$$ in $$CaSO_4$$
3. $$Sn$$ in $$SnO_2$$
4. $$Cd$$ in $$Cd$$
5. $$Co$$ in $$CoCl_2$$

1. Oxidation state of C: +4
2. Oxidation state of S: +6
3. Oxidation state of Sn: +4
4. Oxidation state of Cd: 0
5. Oxidation state of Co: +2

Hint:

Solution:

1. (Oxidation state of C) + (2* Oxidation state of O) = 0

(Oxidation state of C) + (2*-2) = 0

(Oxidation state of C) = +4

1. (Oxidation state of Ca)+(Oxidation state of S)+(4*Oxidation state of O) = 0

(+2) + (Oxidation state of S) + (4*-2) = 0

(Oxidation state of S) = +6

1. (Oxidation state of Sn) + (2* Oxidation state of O) = 0

(Oxidation state of Sn) + (2* -2) = 0

(Oxidation state of Sn) = +4

1. (Oxidation State of Cd) =0
2. (Oxidation state of Co) + (2* Oxidation state of Cl) = 0

(Oxidation state of Co) + (2*-1) = 0

(Oxidation state of Co) = +2

## 3.19

Indicate the oxidation state for each atom in the compound below:

1. $$CaMgO_2$$
2. $$SeF_6$$
3. $$HNO_3$$
4. $$As_2O_3$$
5. $$LiMnO_4$$
6. $$PF_5$$
7. $$H_2SF_6$$
8. $$K_2Cr_2O_7$$
9. $$C_6H_{12}O_6$$
10. $$NH_4OH$$

1. Ca: 2+ Mg: 2+ O=2-
2. Se: 6+ F:1-
3. H: 1+ N: 5+ O: 2-
4. As: 3+ O: 2-
5. Li: 1+ Mn: 7+ O: 2-
6. P: 5+ F: 1-
7. H: 1+ S: 4+ F: 1-
8. K: 1+ Cr: 6+ O: 2-
9. C: 0 H: 1+ O: 2-
10. N: 3- H: 1+ O: 2-

Hint:

Solution

Oxidation state (oxidation number) is a number assigned to an element in a compound that represents the number of electrons lost or gained by the atom in the compound.

a. CaMgO2 is composed of Ca2+, Mg2- and O2- ions. Oxygen always have an oxidation state of -2, Mg also always have an oxidation state of +2, and Calcium also have an oxidation state of +2. In order to verify if the oxidation state is correct we can do this calculation:

=(Ca Oxidation state) + (Mg oxidation state) + (2)(O oxidation state)

=(+2) +(+2) + (2(-2)) =0

The oxidation number of oxygen is multiplied by2 because the compound consists of 2 oxygen. Since the total oxidation number of the compound is 0, the oxidation state of each atom is verified.

b. F has an oxidation state of -1 since F is in group 7A. We are not sure about the oxidation state of Se, therefore we have to do some calculation:

Let the oxidation number of Se be “n”

(n) + (6(-1)) = 0

(n) = +6

A compound that has no charge means has 0 charge

Thus, SeF6 has 0 charge which implies

(oxidation # of Se) +(6(oxidation # of F)) = 0

c. H has an oxidation state of +1 and O has an oxidation state of -2

(Oxidation state of H) + (Oxidation state of N) + (Oxidation state of O (3)) = 0

(+1)+(Oxidation state of N)+(-2(3)) = 0

(Oxidation state of N) = 6-1 = +5

d. O has an oxidation state of -2

(Oxidation state of As (2)) + (Oxidation state of O (3)) = 0

(2* Oxidation state of As) + (-2*3) = 0

Oxidation state of As = 3+

The oxidation state of O is multiplied by 3 and the oxidation state of As is multiplied by 2 by looking at the compound As2O3.

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