EXAMPLE 6.1
Balancing Equations
Balance the following representation of the chemical reaction involved when an airbag deploys.
NaN3  Na + N2
Solution
Sodium atoms are balanced, but the nitrogen atoms are not. For this sort of problem, we will use the concept
of the least common multiple. There are three nitrogen atoms on the left (reactants side) and two on the right
(products side). The least common multiple of 2 and 3 is 6. Therefore, we need three N2 and two NaN3:
2 x 3 = 6 N atoms
3 x 2 = 6 N atoms
2 NaN3  Na + 3 N2
2 Na atoms
(not balanced)
1 Na atoms
We now have two sodium atoms on the left. We can get two on the right by placing the coefficient 2 in front
of Na.
2 NaN3  2 Na + 3 N2
(balanced)
Checking, we count two Na atoms and six N atoms on each side. The equation is balanced.
EXAMPLE 6.1
Balancing Equations continued
Exercise 6.1A
The reaction between hydrogen and nitrogen to give ammonia, called the Haber process, is typically the first
step in the industrial production of nitrogen fertilizers, represented as
H2 + N2  NH3
Balance the equation.
Exercise 6.1B
Iron ores like Fe2O3 are smelted by reaction with carbon to produce metallic iron and carbon dioxide,
represented as
Fe2O3 + C  CO3 + Fe
Balance the equation.
EXAMPLE 6.2
Balancing Equations
When a fuel such as methane is burned in sufficient air, the products are carbon dioxide and water, represented
as
CH4 + O2  CO2 + H2O
(not balanced)
Balance the equation.
Solution
In this equation, oxygen appears in two different products and by itself in 0 2; we leave the oxygen for last
and balance the other two elements first. Carbon is already balanced, with one atom on each side of the
equation. For hydrogen, the least common multiple of 2 and 4 is 4, and so we place the coefficient 2 in front
of H2O to balance hydrogen. Now we have four hydrogen atoms on each side.
CH4 + O2  CO2 + 2 H2O
(not balanced)
Now for the oxygen. There are four oxygen atoms on the right. If we place a 2 in front of O 2 on the left, the
oxygen atoms balance.
CH4 + 2 O2  CO2 + 2 H2O
(balanced)
The equation now has one C atom, four H atoms, and four O atoms on each side; it is balanced.
EXAMPLE 6.2
Balancing Equations continued
Exercise 6.2A
Butane is a common fuel. Its combustion is represented as
C4H10 + O2  CO2 + H2O
Balance the equation.
Exercise 6.2B
Does it take more oxygen (per molecule) to burn butane (see Exercise 6.2A) than it does to burn methane?
How much more CO2 is produced?
EXAMPLE 6.3
Volume Relationships of Gases
What volume of oxygen is required to burn 0.556 L of propane if both gases are measured at the same
temperature and pressure?
C3H8(g) + 5 O2(g)  3 CO2 (g) + 4 H2O(g)
Solution
The coefficients in the equation indicate that each volume of C 3H8(g) requires 5 volumes of O2(g). Thus, we
use 5 L O2(g)/1 L C3H8(g) as the ratio to find the volume of oxygen required.
? L O2(g) = O.556 L C3 H8(g) x
5 L O2(g)
1 L C3 H8(g)
= 2.78 L O2(g)
Exercise 6.3A
Using the equation in Example 6.3, calculate the volume of CO 2(g) produced when 0.492 L of propane is
burned if the two gases are compared at the same temperature and pressure.
Exercise 6.3B
If 10.0 L each of propane and oxygen are combined at the same temperature and pressure, which gas will be
left over after reaction? What volume of that gas will remain?
EXAMPLE 6.4
Calculating Molecular Masses
Calculate (a) the molecular mass of nitrogen dioxide (NO2) an amber colored gas that is a constituent of smog,
and (b) the formula mass of ammonium sulfate [(NH4)2SO4] a fertilizer commonly used by home gardeners.
Solution
a.
We start with the molecular formula: NO2. Then, to determine the molecular mass, we need only to
add the atomic mass of nitrogen to twice the atomic mass of oxygen.
1 x atomic mass of N = 1 x 14.0 u = 14.0 u
2 x atomic mass of O = 2 x 16.0 u = 32.0 u
Formula mass of NO2 = 46.0 u
b.
Using a calculator, we need only write down the final answer, 46.0 u. That is, we have no need to
record the numbers 14.0 and 32.0.
We must make certain that all the atoms in the formula unit are accounted for, which means paying
particular attention to all the subscripts and parentheses in the formula. The “(NH 4)2” means that both
the “N” and the “H4” must be multiplied by 2—that is, the formula indicates a total of two N atoms
and eight H atoms. Combining the atomic masses, we have
2 x atomic mass of N = 2 x 14.0 u
8 x atomic mass of H = 8 x 1.01 u
1 x atomic mass of S = 1 x 32.0 u
4 x atomic mass of O = 4 x 16.0 u
Formula mass of (NH4)2SO4
= 28.0 u
= 8.08 u
= 32.0 u
= 64.0 u
= 132.1 u
EXAMPLE 6.4
Calculating Molecular Masses continued
Exercise 6.4A
Calculate the formula mass of (a) sodium azide (NaN3) used in automobile airbags, and (b) phosphoric acid
(H3PO4).
Exercise 6.4B
Calculate the formula mass of (a) para-dichlorobenzene (C6H4Cl2) used as a moth repellent, and (b) calcium
dihydrogen phosphate [Ca(H2PO4)2], used as a mineral supplement in foods.
EXAMPLE 6.5
Mole-to-Mass Conversions
How many grams of N2 are in 0.400 moles N2?
Solution
The molecular mass of N2 is 2 x 14.0 u = 28.0 u. The molar mass of N2 is therefore 28.0 g/mol. Using the
molar mass as a conversion factor (red), we have
? g N2 = 0.400 mol N2 x
28.0 g N2
1 mol N2
= 11.2 g N2
Exercise 6.5
Calculate the mass, in grams, of (a) 0.0728 mol silicon, (b) 55.5 mol H2O, and (c) 0.0728 mol Ca(H2PO4)2.
EXAMPLE 6.6
Mass-to-Mole Conversions
Calculate the number of moles of Na in a 62.5-g sample of sodium metal.
Solution
The molar mass of Na is 23.0 g/mol. To convert from a mass in grams to an amount in moles, we must use
the inverse of the molar mass as a conversion factor (1 mol Na/23.0 g Na) to get the proper cancellation of
units. When we start with grams, we must have grams in the denominator of our conversion factor (red).
? mol Na = 62.5 g Na x
1 mol Na
23.0 g Na
= 2.72 mol Na
Exercise 6.6
Calculate the amount, in moles, of (a) 3.71 g Fe, (b) 165 g butatne, C4H10, and (c) 0.100 mol Mg(NO3)2.
EXAMPLE 6.7
Density of a Gas at STP
Calculate the density of (a) nitrogen gas and (b) methane (CH4) gas, both at STP.
Solution
a.
The molar ass of N2 gas is 28.0 g/mol. We multiply by the conversion factor 1 mol N 2 = 22.4 L,
arranged to cancel units of moles.
28.0 g N2
1 mol N2
b.
x
1 mol N2
22.4 L N2
= 1.25 g/L
The molar mass of CH4 gas is (1 x 12.0) g/mol + (4 x 1.01) g/mol = 16.0 g/mol. Again we use the
conversion factor 1 mol CH4 = 22.4 L.
16.0 g CH4
1 mol CH4
x
1 mol CH4
22.4 L CH4
= 0.714 g/L
Exercise 6.7A
Calculate the density of He at STP.
Exercise 6.7B
Estimate the density of air at STP (assume 78% N2 and 22% O2) and compare this value to the value of He
you calculated in Exercise 6.7A.
EXAMPLE 6.8
Molar Mass from Gas Densities
The density of diethyl ether vapor is 3.30 g/L. Calculate the molar mass of diethyl ether.
Solution
This time we are given the density in g/L. We want molar mass, which has units of g/mol. Once again
we have the factor 22.4 L = 1 mol diethyl ether. Clearly, we must cancel the unit of liters and obtain the unit
of moles in
the denominator.
3.30 g
22.4 L
x
= 73.9 g/mol
1L
1 mol
Exercise 6.8A
The density of an unknown gas at STP is 2.30 g/L. Calculate its molar mass.
Exercise 6.8B
An unknown gaseous compound contains only hydrogen and carbon and its density is 1.34 g/mol at STP.
What is the formula for the compound?
CONCEPTUAL EXAMPLE 6.9
Molecular, Molar, and Mass
Relationships
Nitrogen monoxide (nitric oxide), an air pollutant discharged by internal combustion engines, combines with
oxygen to form nitrogen dioxide, a yellowish-brown gas that irritates the respiratory system and eyes. The
equation for this reaction is
2 NO + O2  2 NO2
State the molecular, molar, and mass relationships indicated by the equation.
Solution
The molecular and molar relationships can be obtained directly from the equation; no calculation is
necessary. The mass relationship requires a little calculation:
Molecular: Two molecules of NO react with one molecule of O2 to form two molecules of NO2.
Molar: 2 mol of NO reacts with 1 mol of O2 to form 2 mol of NO2.
Mass: 60.0 g of NO (2 mol NO x 30.0 g/mol) reacts with 32.0 g (1 mol O2 x 32.0 g/mol) of O2 to form 92.0
g (2 mol NO2 x 46.0 g/mol) of NO2.
Exercise 6.9
Hydrogen sulfide, a gas that smells like rotten eggs, burns in air to produce sulfur dioxide and water according
to the equation
2 H2S + 3 O2  2 SO2 + H2O
State the molecular, molar, and mass relationships indicated by this equation.
EXAMPLE 6.10
Molar Relationships
When 0.105 mol of propane is burned in a plentiful supply of oxygen, how many moles of oxygen is consumed?
C3H8 + 5 O2  3 CO2 + 4 H2O
Solution
The equation tells us that 5 mol O2 is required to burn 1 mol C3H8. We can write
1 mol C3H8
5 mol O2
where we use the symbol
to mean “is stoichiometrically equivalent to.” From this relationship we can
construct conversion factors to relate moles of oxygen to moles of propane. The possible conversion factors are
1 mol C3H8
5 mol O2
and
5 mol O2
1 mol C3H8
Which one do we use? Only if we multiply the given quantity (0.105 mol C 3H8) by the factor on the right do we
get an answer with the asked-for units (moles of oxygen).
5 mol O2
? mol O2 = 0.105 mol C3H8 x
= 0.525 mol O2
1 mol C3H8
Exercise 6.10
For the combustion of propane in Example 6.10, (a) How many moles of carbon dioxide is formed when
0.529 mol of C3H8 is burned? (b) How many moles of water is produced when 76.2 mol of C3H8 is burned?
(c) How many moles of carbon dioxide is produced when 1.010 mol of O 2 is consumed?
EXAMPLE 6.11
Mass Relationships
Calculate the mass of oxygen needed to react with 10.0 g of carbon in the reaction that forms carbon dioxide.
Solution
Step 1 The balanced equation is
C + O2  CO2
Step 2 The molar masses are 2 x 16.0 = 32.0 g/mol for O2 and 12.0 g/mol for C.
Step 3 We convert the mass of the given substance, carbon, to an amount in moles.
1 mol O2
? mol C = 10.0 g C x
= 0.833 mol C
12.0 g C
Step 4 We use coefficients from the balanced equation equation to establish the stoichiometric factor (red) that
relates the amount of oxygen to that of carbon.
0.833 mol C x
1 mol O2
1 mol C
= 0.833 mol O2
Step 5 We convert from moles of oxygen to grams of oxygen.
32.0 g O2
0.833 mol O2 x
= 26.7 g O2
1 mol O2
EXAMPLE 6.11
Mass Relationships
continued
We can also combine the five steps into a single setup. Note that the units in the denominators of the
conversion factors are chosen so that each cancels the unit in the numerator of the preceding term.
We start here
This converts
g C to mol C
10.0 g C x
1 mol C
12.0 g C
This release
mol C to
mol O2
x
1 mol O2
1 mol C
This converts
mol O2 to
g O2
x
32.0 g O2
1 mol O2
The answer:
the number
and the unit
= 26.6 g O2
(The slightly different answers are due to rounding in the intermediate steps.)
Exercise 6.11A
Calculate the mass of oxygen (O2) needed to react with 0.334 g of nitrogen (N2) in the reaction that forms
nitrogen dioxide.
Exercise 6.11B
Calculate the mass of carbon dioxide formed by burning 775 g of each of (a) methane (CH4) and (b)
butane (C4H10).
EXAMPLE 6.12
Mass Relationships
The decomposition of sodium azide (NaN3) produces sodium metal and nitrogen gas. The gas is used to inflate
automobile airbags. What mass of nitrogen, in grams, can be made from 60.0 g of sodium azide?
Solution
1.
We start by writing and balancing the chemical equation, which shows that 2 mol NaN 3 produces 2 mol
Na and 3 mol N2.
2 NaN3  2 Na + 3 N2
2.
3.
The molar mass of NaN3 is 23.0 g/mol + (3 x 14.0) g/mol = 65.0 g/mol and the molar mass of N2 is (2 x
14.0) g/mol = 28.0 g/mol.
We convert the mass of the given substance, sodium azide, to an amount in moles.
60.0 g NaN3 x
4.
1 mol NaN3
= 0.923 mol NaN3
65.0 g
NaN
We use coefficients from the balanced equation to3establish the stoichiometric factor that relates the
amount of nitrogen gas to that of sodium azide.
3 mol N2
0.923 mol NaN3 x
= 1.38 mol N2
2 mol NaN3
EXAMPLE 6.12
5.
Mass Relationships continued
We convert from moles of nitrogen gas to grams of nitrogen gas.
28.0 g N2
1.38 mol N2 x
= 38.6 g N2
1 mol N2
As is usually the case, all the steps just outlined can be combined into a single setup.
1 mol NaN3 3 mol NaN3
28.0 g N2
60.0 g NaN3 x
x
x
= 38.8 g NaN3
65.0 g
2 mol NaN3
1 mol N2
NaN3
Notice that the units of the numerator in one stoichiometric factor are the units in the denominator
of the next stoichiometric factor. In this way, the correct cancellation of units occurs and the units of
the final numerator are the units of your answer.
Exercise 6.12A
Ammonia reacts with phosphoric acid (H3PO4) to form ammonium phosphate [(NH4)3PO4]. What mass in
grams of ammonia is needed to react completely with 74.8 g of phosphoric acid?
Exercise 6.12B
a.
b.
The decomposition of potassium chlorate (KCIO3) produces potassium chloride (KCI) and O2 gas.
What mass in grams of oxygen can be made from 2.47 g of potassium chlorate?
Phosphorus reacts with oxygen to form tetraphosphorus decoxide. The equation is
P4 + O2  P4O10
(not balanced)
What mass in grams of tetraphosphorus decoxide can be made from 3.50 g of phosphorus?
EXAMPLE 6.13
Boyle's Law: Pressure-Volume Relationships
A gas is enclosed in a cylinder fitted with a piston. The volume of the gas is 2.00 L at 0.524 atm. The piston is
moved to increase the gas pressure to 5.15 atm. Which of the following is a reasonable value for the volume of
the gas at the greater pressure?
0.20 L
0.40 L
1.00 L
16.0 L
Solution
The pressure increase from 0.524 atm to 5.15 atm is almost tenfold. The volume should drop to about one-tenth
of the initial value. We estimate a volume of 0.20 L. (The calculated value is 0.203 L.)
Exercise 6.13
A gas is enclosed in a 10.2-L tank at 1208 mmHg. (The mmHg is a pressure unit; 760 mmHg = 1 atm.) Which
of the following is a reasonable value for the pressure when the gas is transferred to a 30.0-L tank?
300 mmHg
400 mmHg
3,600 mmHg
12,000 mmHg
Boyle’s Law: Pressure-Volume Relationships
EXAMPLE 6.14
A cylinder of oxygen has a volume of 2.25 L. The pressure of the gas is 1470 pounds per square inch (psi) at
20 °C. What volume will the oxygen occupy at standard atmospheric pressure (14.7 psi) assuming no
temperature change?
Solution
We find it helpful to first separate the initial from the final condition.
Initial
Final
P1 = 1470 psi
P2 = 14.7 psi
V1 = 2.25 L
V2 = ?
Change
The pressure goes
down, therefore, the
volume goes up.
Then use the equation V1P1 = V2P2 and solve for the desired volume or pressure. In this case, we solve for V2.
V2 =
V2 =
V1P1
P2
2.25 L x 1470 psi
14.7 psi
= 225 L
Because the final pressure (14.7 psi) is less than the initial pressure (1470 psi), we expect the final volume
(225 L) to be larger than the original volume (2.25 L), and we see that it is.
EXAMPLE 6.14
Boyle’s Law: Pressure-Volume Relationships
continued
Exercise 6.14A
A sample of air occupies 73.3 mL at 98.7 atm and 0 ºC. What volume will the air occupy at 4.02 atm and 0 ºC?
Exercise 6.14B
A sample of helium occupies 535 mL at 988 mmHg and 25 °C. If the sample is transferred to a 1.05-L flask
at 25 °C, what will be the gas pressure in the flask?
Charles’s Law: Temperature-Volume
Relationships
EXAMPLE 6.15
A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L. What would its volume be (a) in a
hot room where the temperature is 47 °C, and (b) outdoors, where the temperature is –23 ºC? (Assume no
change in pressure in either case.)
Solution
First, and most important, convert all temperatures to the Kelvin scale:
T(K) = t(°C) + 273
The initial temperature (T1) in each case is (27 + 273) = 300 K and the final temperatures are (a) (47 + 273) =
320 K and (b) (–23 + 273) = 250 K.
a.
We start by separating the initial from the final condition.
Initial
t1 = 27 °C
T1 = 300 K
V1 = 2.00 L
Solving the equation
Final
t2 = 47 °C
T2 = 320 K
V2 = ?
V1
T1
Change
V2
=
T2
Charles’s Law: Temperature-Volume
Relationships continued
EXAMPLE 6.15
for V2, we have
V2 =
V2 =
V1T2
T1
2.00 L x 320 K
300 K
= 2.13 L
As expected, because the temperature increases, the volume must also increase.
b.
We have the same initial conditions as in (a), but different final conditions.
Initial
Final
t1 = 27 °C
t2 = -23 °C
T1 = 300 K
T2 = 250 K
V1 = 2.00 L
V2 = ?
Change
Again using Charles’s law, we solve the equation for V2:
V2 =
V2 =
V1T2
T1
2.00 L x 250 K
= 1.67 L
300 K
As we expected, the volume decreased because the temperature decreased.
EXAMPLE 6.15
Charles’s Law: Temperature-Volume
Relationships continued
Exercise 6.15A
a.
b.
A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. What volume will this sample occupy
at 150 °C? (Assume no change in pressure.)
A sample of hydrogen occupies 692 L at 602 °C. If the pressure is held constant, what volume will the
gas occupy after being cooled to 23 °C?
Exercise 6.15B
At what Celsius temperature will the initial volume of oxygen in Exercise 6.15A occupy 0.750 L? (Assume no
change in pressure.)
Ideal Gas Law
EXAMPLE 6.16
Use the ideal gas law to calculate (a) the volume occupied by 1.00 mol of nitrogen gas at 244 K and 1.00 atm
pressure, and (b) the pressure exerted by 0.500 mol of oxygen in a 15.0-L container at 303 K.
Solution
a.
We start by solving the ideal gas equation for V.
V=
V=
b.
nRT
P
1.00 mol
1.00 atm
x
0.0821 L • atm
mol • K
x 244 K = 20.0 L
Here we solve the ideal gas equation for P.
P=
P=
nRT
V
0.500 mol
15.0 L
x
0.0821 L • atm
mol • K
x 303 K = 0.829 atm
Exercise 6.16A
Determine (a) the pressure exerted by 0.0330 mol of oxygen in an 18.0-L container at 313 K, and (b) the
volume occupied by 0.200 mol of nitrogen gas at 298 K and 0.980 atm.
EXAMPLE 6.16
Ideal Gas Law continued
Exercise 6.16B
Determine the volume of nitrogen gas produced from the decomposition of 130 g sodium azide (about the
amount in a typical automobile airbag) at 25 °C and 1 atm.
EXAMPLE 6.17
Solution Concentration: Molarity
Calculate the molarity of a solution made by dissolving 3.50 mol of NaCl in enough water to produce 2.00 L
of solution.
Solution
Molarity (M) =
moles of solute
liters of solution
=
3.50 mol NaCl
2.00 L solution
= 1.75 M NaCl
We read 1.75 M NaCl as “1.75 molar NaCl.”
Exercise 6.17A
Calculate the molarity of a solution that has 0.0500 mol of NH3 in 5.7 L of solution.
Exercise 6.17B
Calculate the molarity of a solution made by dissolving 0.750 mol of H 3PO4 in enough water to produce
775 mL of solution.
Solution Concentration: Molarity
EXAMPLE 6.18
What is the molarity of a solution in which 333 g of potassium hydrogen carbonate is dissolved in enough water
to make 10.0 L of solution?
Solution
First, we must convert grams of KHCO3 to moles of KHCO3
333 g KHCO3 x
1 mol KHCO3
100.1 g KHCO3
= 3.33 mol KHCO3
Now use this value as the numerator in the defining equation for molarity. The solution volume, 10.0 L, is
the denominator.
Molarity =
3.33 mol KHCO3
10.0 L solution
= 0.333 M KHCO3
Exercise 6.18
Calculate the molarity of each of the following solutions.
a.
18.0 mol of H2SO4 in 2.00 L of solution
b.
3.00 mol of KI in 2.39 L of solution
c.
0.206 mol of HF in 752 mL of solution (HF is used for etching glass.)
EXAMPLE 6.19
Solution Preparation: Molarity
How many grams of NaCl is required to prepare 0.500 L of typical over-the-counter saline solution
(about 0.15 M NaCl)?
Solution
First we use the molarity as a conversion factor to calculate moles of NaCl.
0.500 L solution x
0.15 mol NaCl
1 L solution
= 0.075 mol NaCl
Then we use the molar mass to calculate the grams of NaCl.
0.75 mol NaCl x
58.4 g NaCl
1 mol NaCl
= 44 g NaCl
Exercise 6.19A
What mass in grams of potassium hydroxide is required to prepare 2.00 L of 6.00 M KOH?
Exercise 6.19B
What mass in grams of potassium hydroxide is required to prepare 100.0 mL of 1.00 M KOH?
EXAMPLE 6.20
Moles from Molarity and Volume
Concentrated hydrochloric acid has a concentration of 12.0 M HCl. How many milliliters of this solution
would one need to get 0.425 mol of HCl?
Solution
Liters of HCl solution =
=
moles of solute
molarity
0.425 mol HCl
12.0 mol HCl/L
=
0.425 mol HCl
12.0 M HCl
= 0.0354 L
We would need 0.0354 L (35.4 mL) of the solution to have 0.425 mol. Remember that molarity is moles per
liter of solution, not per liter of solvent.
Exercise 6.20A
What volume in milliliters of 15.0 M aqueous ammonia (NH3) solution do you need to get 0.445 mol of NH3?
Exercise 6.20B
What mass in grams of HNO3 is in 500 mL of rain that has a concentration of 2.0 x 10–5 M HNO3?
EXAMPLE 6.21
Percent by Volume
Two-stroke engines use a mixture of 120 mL of oil dissolved in enough gasoline to make of 4.0 liters of fuel.
What is the percent by volume of oil in this mixture?
Solution
Percent by volumes =
120 mL oil
4000 mL solution
x 100% = 3.0%
Exercise 6.21A
What is the volume percent of ethanol in a solution that has 58.0 mL water in 625 mL of an ethanol–water
solution?
Exercise 6.21B
Assume that the volumes are additive, and determine the volume percent toluene (C 6H5CH3) in a solution
made by mixing 40.0 mL of toluene with 75.0 mL of benzene (C 6H6).
EXAMPLE 6.22
Solution Preparation: Percent by Volume
Describe how to make 775 mL of vinegar (about a 5.0% by volume solution of acetic acid in water).
Solution
We begin by rearranging the equation for percent by volume to solve for volume of solute.
Percent by volume x volume of solution
Volume of solute =
100%
Substituting, we have
=
5.0% x 775 mL
100%
= 39 mL
Take 39 mL of acetic acid and add enough water to make 775 mL of solution. Notice that we don’t simply add
775 mL of water, because the final volume of solution must be 775 mL.
Exercise 6.22A
Describe how to prepare 450 mL of an aqueous solution that is 70.0% isopropyl alcohol by volume.
Exercise 6.22B
Describe how you would prepare exactly 2.00 L of an aqueous solution that is 9.77% acetic acid by volume.
EXAMPLE 6.23
Percent by Mass
What is the percent by mass of a solution of 25.5 g of NaCl dissolved in 425 g (425 mL) of water?
Solution
Use these values in the above percent-by-mass equation:
25.5 g NaCl
Percent by mass =
x 100% = 5.66% NaCl
(25.5 + 425) g solution
Exercise 6.23A
Hydrogen peroxide solutions for home use are 3.0% by mass solutions of H 2O2 in water. What is the percent
by mass of a solution of 9.40 g of H2O2 dissolved in 335 g (335 mL) of water?
Exercise 6.23B
Sodium hydroxide (NaOH, lye) is used to make soap and is very soluble in water. What is the percent by mass
of a solution that contains 1.00 kg of NaOH dissolved in 950 mL of water?
EXAMPLE 6.24
Solution Preparation: Percent by Mass
Describe how to make 430 g of an aqueous solution that is 4.85% by mass NaNO 3.
Solution
We begin by rearranging the equation for percent by mass to solve for mass of solute.
percent by mass x mass of solution
Mass of solute =
100 %
Substituting, we have
=
4.85% x 430 g
100 %
= 20.9 g
Take 20.9 g of NaNO3 and add enough water to make 430 g of solution.
Exercise 6.24A
Describe how you would prepare 125 g of an aqueous solution that is 4.50% glucose by mass.
Exercise 6.24B
Describe how you would prepare 1750 g of isotonic saline, a commonly used intravenous (IV) solution that is
0.89% sodium chloride by mass.