Homework Solutions #1

Q1

Complete each calculation with the correct number of significant figures

1. \(\dfrac{7.68 \times 8.009}{4.1}\phantom{\dfrac{\Big(}{\Big(}}\)
2. \(\dfrac{6.89-5.452}{1.596+4.7121}\phantom{\dfrac{\Big(}{\Big(}}\)
3. \(\dfrac{3.668\times10^{-7}+ 8.3469\times10^{-7}}{3.7348\times10^{-10}}\phantom{\dfrac{\Big(}{\Big(}} \)

Answer

1. 15
2. 0.23
3. 3217

Hint

• http://chemwiki.ucdavis.edu/Under_Construction/SARIS/Liini/Towson/Significant_Figures

Solution

It is easier to take into account the correct number of significant figures after the calculations are completed to avoid a rounding error. Once you calculate the answer, determine which number will limit your answer and round accordingly.

Q2

Calculate how many grams of mixture are there if 250 mL of ethanol is mixed with 500 mL of benzene.

Substance	Density @ 20^oC, 1 atm (g/cm^3)
Ethanol	0.789
Benzene	0.880

Answer

• 637 g.

Hint

• http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Bulk_Properties/Density

Solution

[(250. mL ethanol / 1) * (1 cm³ethanol / 1 mL ethanol) * (0.789 g ethanol / 1 cm³ ethanol)] + [500. mL benzene / 1) * (1 cm³benzene / 1 mL benzene) * (0.880 g benzene / 1 cm³ benzene)] = 637 g (3 sig. figs)

Q3

Give the ISOTOPIC symbol for the following ions:

1. Ion with 73 Protons, 75 Electrons, and 107 Neutrons.
2. Ion with 28 Protons, 25 Electrons, and 31 Neutrons
3. Ion with 37 Protons, 37 Electrons, and 49 Neutrons
4. Ion with 7 Protons, 7 Electrons, and 7 Neutrons

Answer

1. \(\ce{^{180}_{73}Ta^{2-}}\)
2. \(\ce{^{59}_{28}Ni^{3+}}\)
3. \(\ce{^{86}_{37}Rb}\)
4. \(\ce{^{14}_{7}N}\)

Hint

• http://chemwiki.ucdavis.edu/Physical_Chemistry/Atomic_Theory/The_Atom/Sub-Atomic_Particles

Solution

When naming an ion, the element can be identified by the number of protons. The atomic weight is calculated by the addition of protons to neutrons. Next, charge is calculated by comparing the quantity of Protons to Neutrons, which Protons provide positive charge and Electrons a negative charge.

Q4

Label the following elements as halogens, alkali metals, alkaline earth metals, noble gases, or transition metals.

1. Lithium,
2. Chlorine
3. Argon
4. Krypton
5. Calcium
6. Barium
7. Cesium
8. Fluorine
9. Radon
10. Tungsten

Answer

1. Alkali metals – Lithium, Cesium
2. Alkaline- Earth – Calcium, Barium
3. Halogens – Fluorine, Chlorine
4. Noble gases – Argon, Krypton, Radon
5. Transition metal – Tungsten

Hint

• Use a periodic table of elements to see how groups of elements are classified.
• chemwiki.ucdavis.edu/Wikitext...Periodic_Table

Solution

Alkali metals are found in the first column of the periodic table. Alkaline – Earth metals are found in Column 2 of the periodic table. Halogens are found in Column 17, Noble gases are found in Column 18, and transition metals are found in column 3-12.

Q5

An element consists of the following isotopes. What is the average atomic mass and the name of the element?

Percent natural abundance	Mass (u)
90.48%	19.9924
0.27%	20.9938
9.25%	21.9914

Answer

• 20.18 amu; Ne.

Hint

• http://chemwiki.ucdavis.edu/Physical_Chemistry/Atomic_Theory/Atomic_Mass

Solution

A = (0.9048*19.9924) + (0.0027*20.9938) + (0.0925*21.9914) = 18.09 + 0.057 + 2.03 = 20.18 amu.

From the periodic table, the element is neon (Ne).

Q6

A compound, made only from mercury and sulfur, is 86.2% \(Hg\) by mass. The molar mass of this compound is 465 g/mol. Determine the empirical formula and the molecular formula of this compound.

Answer

• Empirical Formula: \(HgS\);
• Molecular Formula: \(Hg_2S_2\).

Hint

• http://chemwiki.ucdavis.edu/Under_Construction/SARIS/Liini/Towson/CONCENTRATION_UNITS/DETERMINING_EMPIRICAL_AND_MOLECULAR_FORMULAE

Solution

The compound is 86.2% S by mass. Therefore, there are 86.2 gram of mercury in 100. gram of the compound.

g S = 100. g – 86.2 g = 13.8 g S.

(86.2 g Hg) * (1 mol Hg / 200.59 g Hg) = 0.430 mol Hg.

(13.8 g S) * (1 mol S / 32.066 g S) = 0.430 mol S.

ratio of Hg : S = 1:1.

So, the empirical formula is \(HgS\)

Empirical mass of \(HgS\) = 200.59 g/mol + 32.066 g/mol = 232.66 g/mol

Multiplication factor = molecular mass / empirical mass = (465 g/mol) / (232.66 g/mol) = 1.999 ≈ 2.

Therefore, molecular formula = (multiplication factor) * (empirical formula) = 2 * HgS = \(Hg_2S_2\).

Q7

A compound is a common first-aid antiseptic. The products of its combustion reaction in the presence of oxygen gas are water and carbon dioxide. This compound is composed of carbon, hydrogen and oxygen. If 9.214 g of this compound produces 17.604 g carbon dioxide and 10.812 g water, what is the empirical formula of this compound? Hint: Some oxygen atoms of the products may come from oxygen gas.

Answer

• \(C_2H_6O\)

Hint

• All carbon and hydrogen atoms of the products come from this compound. What is combustion reaction?
• http://chemwiki.ucdavis.edu/Analytic...ical_Reactions

Solution

(17.604+10.812)-9.214=19.200g oxygen gas

19.200g/16.00(g/mol)=1.200mol O atom

17.604g/44.01(g/mol)=0.4000mol carbon dioxide, 0.4000mol carbon, 0.8000mol O atom

10.812g/18.02(g/mol)=0.6000mol water, 1.200 mol hydrogen, 0.6000mol O atom

Ratio of C, H and O of this compound: 0.4000 C: 1.200 H : (0.8000+0.6000-1.2000) O = 0.40 C:1.2 H:0.20 O = 2 C : 6 H : 1 O

Empirical Formula: \(C_2H_6O\)

Q8

Assign the oxidation state of each atom for the following compounds.

1. \(Al\)
2. \(PbSO_4\)
3. \(H_2SO_3\)
4. \(NaHCO_3\)
5. \(H_2O_2\)
6. \(Ne\)
7. \(MgCO_3\)
8. \(Na_2Cr_2O_7\)

Answer

1. 0
2. Pb: +2 , S: +6, O: -2
3. H:+1, S: +4, O: -2
4. Na: +, H: +, C: +4, O:-2
5. H:+, O: -1
6. Ne: 0
7. Mg +2. C+4,O-2
8. Na+, Cr +6, O-2

Hint

• http://chemwiki.ucdavis.edu/Under_Construction/SARIS/Liini/Towson/Determining_Oxidation_States

Solution

• Oxidation state of atoms of all metals, pure substances and noble gases are 0.
• The sum of the oxidation state of a compound is 0.