6.4: Counting Molecules by the Gram
- Page ID
- 369350
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Learning Objectives
- Define molecular mass and formula mass.
- Perform conversions between mass and moles of a compound.
- Perform conversions between mass and number of particles.
Molecular and Formula Masses
The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example \(\PageIndex{1}\).
Example \(\PageIndex{1}\): Ethanol
Calculate the molecular mass of ethanol, whose condensed structural formula is \(\ce{CH_3CH_2OH}\). Among its many uses, ethanol is a fuel for internal combustion engines
Solution
Steps for Problem Solving | Calculate the molecular mass of ethanol, whose condensed structural formula is \(\ce{CH_3CH_2OH}\) |
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Identify the "given"information and what the problem is asking you to "find." | Given: Ethanol molecule (CH3CH2OH) Find: molecular mass |
Determine the number of atoms of each element in the molecule. |
The molecular formula of ethanol may be written in three different ways:
All show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom. |
Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element. |
1 C atom = 12.01 amu 1 H atom = 1.008 amu 1 O atom = 16.00 amu |
Add the masses together to obtain the molecular mass. |
2C: (2 atoms)(12.011amu/atom) = 24.02 amu 6H: (6 atoms)(1.0079amu/atom) = 6.047 amu +1O: (1 atoms)(15.9994amu/atom) =16.00 amu C2H6O : molecular mass of ethanol = 46.07 amu |
Exercise \(\PageIndex{1}\): Freon
Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, which has a condensed structural formula of \(\ce{CCl3F}\). Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:
![158008297991312327.png](https://chem.libretexts.org/@api/deki/files/260371/158008297991312327.png?revision=1&size=bestfit&width=238&height=201)
- Answer
- 137.37 amu
Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units.
Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
Example \(\PageIndex{2}\): Calcium Phosphate
Calculate the formula mass of \(\ce{Ca3(PO4)2}\), commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.
Solution
Steps for Problem Solving | Calculate the formula mass of \(\ce{Ca3(PO4)2}\), commonly called calcium phosphate. |
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Identify the "given" information and what the problem is asking you to "find." | Given: Calcium phosphate [Ca3(PO4)2] formula unit Find: formula mass |
Determine the number of atoms of each element in the molecule. |
|
Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element. |
1 Ca atom = 40.08 amu 1 P atom = 30.97 amu 1 O atom = 16.00 amu |
Add together the masses to give the formula mass. |
3Ca: (3 atoms) (40.08 amu/atom)=120.2 amu 2P: (2 atoms) (30.97 amu/atom)=61.95 amu + 8O: (8 atoms)(16.00 amu/atom)=128.0 amu Formula mass of Ca3(PO4)2=310.2 amu |
Exercise \(\PageIndex{2}\): Silicon Nitride
Calculate the formula mass of \(\ce{Si3N4}\), commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.
![Silicon nitride (Si3N4) bearing parts.](https://chem.libretexts.org/@api/deki/files/114092/Si3N4bearings.jpg?revision=1&size=bestfit&width=266&height=229)
- Answer
- 140.29 amu
Molar Masses of Compounds
Similar to how the atomic mass of an element can be related to the molar mass of an element, \[12.01\ amu\ carbon = 12.01\ g/mole \ carbon\] the molecular or formula mass of a compound can be related to its molar mass. \[18.02\ amu\ water = 18.02\ g/mole\ water\]
The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
The periodic table lists the atomic mass of carbon as 12.01 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.01 g/mol:
Substance (formula) | Basic Unit | Atomic, Molecular, or Formula Mass (amu) | Molar Mass (g/mol) |
---|---|---|---|
carbon (C) | atom | 12.01 (atomic mass) | 12.01 |
ethanol (C2H5OH) | molecule | 46.06 (molecular mass) | 46.06 |
calcium phosphate [Ca3(PO4)2] | formula unit | 310.17 (formula mass) | 310.17 |
Converting Between Grams and Moles of a Compound
The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride \(\left( \ce{CaCl_2} \right)\). Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. Dimensional analysis will allow you to calculate the mass of \(\ce{CaCl_2}\) that you should measure as shown in Example \(\PageIndex{3}\).
Example \(\PageIndex{3}\): Calcium Chloride
Calculate the mass of 3.00 moles of calcium chloride (CaCl2).
![Calcium chloride (anhydrous).](https://chem.libretexts.org/@api/deki/files/78370/CK12_Screenshot_10-4-1.png?revision=1&size=bestfit&width=199&height=195)
Solution
Steps for Problem Solving | Calculate the mass of 3.00 moles of calcium chloride (CaCl2). |
---|---|
Identify the "given" information and what the problem is asking you to "find." | Given: 3.00 moles of CaCl2 Find: g CaCl2 |
List other known quantities. | 1 mol CaCl2 = 110.98 g CaCl2 |
Prepare a concept map and use the proper conversion factor. |
|
Cancel units and calculate. | \(3.00 \: \cancel{\text{mol} \: \ce{CaCl_2}} \times \dfrac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \cancel{\text{mol} \: \ce{CaCl_2}}} = 333 \: \text{g} \: \ce{CaCl_2}\) |
Think about your result. |
Exercise \(\PageIndex{3}\): Calcium Oxide
What is the mass of \(7.50 \: \text{mol}\) of (calcium oxide) \(\ce{CaO}\)?
- Answer
- 420.60 g
Example \(\PageIndex{4}\): Water
How many moles are present in 108 grams of water?
Solution
Steps for Problem Solving | How many moles are present in 108 grams of water? |
---|---|
Identify the "given" information and what the problem is asking you to "find." | Given: 108 g H2O Find: mol H2O |
List other known quantities. | \(1 \: \text{mol} \: \ce{H_2O} = 18.02 \: \text{g}\) H2O |
Prepare a concept map and use the proper conversion factor. |
|
Cancel units and calculate. | \(108 \: \cancel{\text{g} \: \ce{H_2O}} \times \dfrac{1 \: \text{mol} \: \ce{H_2O}}{18.02 \: \cancel{\text{g} \: \ce{H_2O}}} = 5.99 \: \text{mol} \: \ce{H_2O}\) |
Think about your result. |
Exercise \(\PageIndex{4}\): Nitrogen Gas
What is the mass of \(7.50 \: \text{mol}\) of Nitrogen gas \(\ce{N2}\)?
- Answer
- 210 g
Conversions Between Mass and Number of Particles
In "Conversions Between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to moles. To convert from mass to number of particles or vice-versa, it will first require a conversion to moles as shown in Figure \(\PageIndex{1}\) and Example \(\PageIndex{5}\).
Example \(\PageIndex{5}\): Chlorine
How many molecules is \(20.0 \: \text{g}\) of chlorine gas, \(\ce{Cl_2}\)?
Solution
Steps for Problem Solving | How many molecules is \(20.0 \: \text{g}\) of chlorine gas, \(\ce{Cl_2}\)? |
---|---|
Identify the "given" information and what the problem is asking you to "find." | Given: 20.0 g Cl2 Find: # Cl2 molecules |
List other known quantities. | 1 mol Cl2 = 70.90 g Cl2, 1mol Cl2 = 6.022 x 1023 Cl2 molecules |
Prepare a concept map and use the proper conversion factor. |
|
Cancel units and calculate. | \(20.0 \: \cancel{\text{g} \: \ce{Cl_2}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{Cl_2}}}{70.90 \: \cancel{\text{g} \: \ce{Cl_2}}} \times \dfrac{6.02 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}}{1 \: \cancel{\text{mol} \: \ce{Cl_2}}} = 1.70 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}\) |
Think about your result. | Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro's number. |
Exercise \(\PageIndex{5}\): Calcium Chloride
How many formula units are in 25.0 g of CaCl2 ?
- Answer
- 1.36 x 1023 CaCl2 formula units
Summary
- Calculations for formula mass and molecular mass are described.
- Calculations involving conversions between moles of a material and the mass of that material are described.
- Calculations are illustrated for conversions between mass and number of particles.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
Henry Agnew (UC Davis)