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# 6.4: Counting Molecules by the Gram

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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 $$\ce{CH_3CH_2OH}$$
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:

• CH3CH2OH (which illustrates the presence of an ethyl group
• CH3CH2−, and an −OH group)
• C2H5OH, and C2H6O;

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 = 12.011 amu

1 H atom = 1.0079 amu

1 O atom = 15.9994 amu

Add the masses together to obtain the molecular mass.

2C: (2 atoms)(12.011amu/atom) = 24.022 amu

6H: (6 atoms)(1.0079amu/atom) = 6.0474amu

+1O: (1 atoms)(15.9994amu/atom) =15.9994amu

C2H6O : molecular mass of ethanol = 46.069amu

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: Figure $$\PageIndex{1}$$: Molecular structure of freon-11, $$\ce{CCl_3F}$$.
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.
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.

• The empirical formula—Ca3(PO4)2—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca2+ ions and two PO43− ions.
• The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.

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 = 40.078 amu

1 P atom = 30.973761 amu

1 O atom = 15.9994 amu

Add together the masses to give the formula mass.

3Ca: (3 atoms) (40.078 amu/atom)=120.234amu

2P: (2 atoms) (30.973761amu/atom)=61.947522amu

+ 8O: (8 atoms)(15.9994amu/atom)=127.9952amu

Formula mass of Ca3(PO4)2=310.177amu

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. Figure $$\PageIndex{2}$$: $$\ce{Si_3N_4}$$ bearing parts. (Public Domain; David W. Richerson and Douglas W. Freitag; Oak Ridge National Laboratory).
140.29 amu

## Molar Mass

The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 1023 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.

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 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 1023 carbon atoms—is therefore 12.011 g/mol:

Table $$\PageIndex{1}$$: Molar Mass of Select Substances
Substance (formula) Basic Unit Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol)
carbon (C) atom 12.011 (atomic mass) 12.011
ethanol (C2H5OH) molecule 46.069 (molecular mass) 46.069
calcium phosphate [Ca3(PO4)2] formula unit 310.177 (formula mass) 310.177

## 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). Figure $$\PageIndex{3}$$: Calcium chloride is used as a drying agent and as a road deicer.

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. $$3.00 \: \cancel{\text{mol} \: \ce{CaCl_2}} \times \dfrac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \cancel{\text{mol} \: \ce{CaCl_2}}} = 333 \: \text{g} \: \ce{CaCl_2}$$

Exercise $$\PageIndex{3}$$: Calcium Oxide

What is the mass of $$7.50 \: \text{mol}$$ of (calcium oxide) $$\ce{CaO}$$?

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. $$108 \: \cancel{\text{g} \: \ce{H_2O}} \times \dfrac{1 \: \text{mol} \: \ce{H_2O}}{18.02 \: \cancel{\text{g} \: \ce{H_2O}}} = 5.99 \: \text{mol} \: \ce{H_2O}$$

Exercise $$\PageIndex{4}$$: Nitrogen Gas

What is the mass of $$7.50 \: \text{mol}$$ of Nitrogen gas $$\ce{N2}$$?

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 $$\PageIndex{5}$$. Figure $$\PageIndex{4}$$: Conversion from number of particles to mass, or from mass to number of particles, requires two steps.

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 ?

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.