Calculations involving molality, molarity, density, mass percent, mole ratio
Ten Examples

Problems 1 - 10

Problems 11 - 25

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Assume in all cases that water is the solvent. The molar mass of water is 18.015 g/mol and the molar mass of sulfuric acid is 98.078 g/mol.

Example #1: Given a density of 1.836 g/mL and a mass percent of H2SO4 of 96.00%, find the molarity, molality, and mole fraction.

Solution:

1) Assume that a volume of 1.000 L of the solution is present. Determine the total mass of the solution:

(1.836 g/mL) (1000. mL) = 1836 g

2) Determine the mass of each component of the solution:

H2SO4 ---> (1836 g) (0.9600) = 1762.56 g
H2O ---> 1836 g minus 1762.56 g = 73.44 g

3) Determine the moles of each component of the solution:

H2SO4 ---> 1762.56 g / 98.078 g/mol = 17.9710 mol
H2O ---> 73.44 g / 18.015 g/mol = 4.0766 mol

4) Determine the mole fraction of each component of the solution:

17.9710 mol + 4.0766 mol = 22.0476 mol

H2SO4 ---> 17.9710 mol / 22.0476 mol = 0.8151
H2O ---> 4.0766 mol / 22.0476 mol = 0.1849

Often the last mole fraction is obtained by subtraction:

H2O ---> 1 - 0.8151 = 0.1849 <--- The 1 is NOT one significant figure.

5) Determine the molarity of the solution:

17.9710 mol / 1.000 L = 17.97 M (to four sig figs)

6) Determine the molality of the solution:

73.44 g = 0.07344 kg

17.9710 mol / 0.07344 kg = 244.7 m

Scorpio reacts.


At example #5 below, I repeat example #1, but I solve it from a different starting point. I assume 100.0 g of the solution is present and I use the weight percent to determine the mass of each solution component. You are invited to try the rest of the solution on your own before looking at my explanation.


Example #2: Given a density of 1.769 g/mL, and a H2SO4 mole fraction of 0.5000, find the molality, molarity, and mass percent.

Solution:

1) We will use a mole fraction of 0.5000 to mean 0.5000 mole is present in a total of 1.0000 mole of solution. Determine the mass of each mole fraction:

H2SO4 ---> (0.5000 mol) (98.078 g/mol) = 49.039 g
H2O ---> (0.5000 mol) (18.015 g/mol) = 9.0075 g

Comment: A mole fraction of 0.50 could mean 1.0 mol of one component in 2.0 total moles. Using 0.5 and 1 is the simplest meaning of a mole fraction of 0.5. We could have used any paring of numbers that gives a mole fraction of 0.5. The final answers would be the same, but the numbers in the calculations would be different.

2) Determine the mass percent of each component:

49.039 g + 9.0075 g = 58.0465 g

H2SO4 ---> 49.039 g / 58.0465 g = 84.48%
H2O ---> 9.0075 g / 58.0465 g = 15.52%

Often the last mass percent is obtained by subtraction:

H2O ---> 100.00 - 84.48 = 15.52

3) Determine the molality:

9.0075 g = 0.0090075 kg

0.5000 mol / 0.0090075 kg = 55.51 m

4) Determine the molarity:

58.0465 g / 1.769 g/mL = 32.81 mL = 0.03281 L

0.5000 mol / 0.03281 L = 15.24 M


Example #3: Given a density of 1.059 g/mL and a H2SO4 molarity of 1.000 M, find the molality, mole fraction, and mass percent.

Solution:

1) Assume 1.0000 L of the solution is present. Determine the mass of the solution:

(1.059 g/mL) (1000.0 mL) = 1059 g

2) Determine the mass percent of each component:

H2SO4 ---> 98.078 g (remember, it's a 1 M solution)
H2O ---> 1059 g - 98.078 g = 960.922 g

H2SO4 ---> 100 - 90.74 = 9.26%
H2O ---> 960.922 g / 1059 g = 90.74%

Note that I calculated the larger value by division and the smaller value by subtraction.

3) Determine the mole fraction:

H2SO4 ---> 98.078 g / 98.078 g/mol = 1.000 mol (or just remember it's a 1 M solution)
H2O ---> 960.922 g / 18.015 g/mol = 53.349 mol

53.349 mol + 1.000 mol = 54.349 mol

H2SO4 ---> 1 - 0.9816 = 0.0184
H2O ---> 53.349 mol / 54.349 mol = 0.9816

4) Determine the molality:

960.922 g = 0.960922 kg

1.000 mol / 0.960922 kg = 1.041 m


Example #4: Given a density 1.122 g/mL and a H2SO4 molality of 4.500 m, find the molarity, mole fraction and mass percent.

Solution:

1) The given molality means 4.500 mol dissolved in 1.000 kg of water. Determine the mass of each component:

H2SO4 ---> (4.500 mol) (98.078 g/mol) = 441.351 g
H2O ---> 1.000 kg = 1000. g

2) Determine mass percentages:

1000. g + 441.351 g = 1441.351 g (total mass of the solution)

H2O ---> 1000. g / 1441.351 g = 69.38%
H2SO4 ---> 100 - 69.38 = 30.62%

3) Determine mole fraction:

H2SO4 ---> 4.500 mol
H2O ---> 1000. g / 18.015 g/mol = 55.509 mol

55.509 mol + 4.500 mol = 60.009 mol

H2SO4 ---> 1 - 0.9250 = 0.0750
H2O ---> 55.509 mol / 60.009 mol = 0.9250

4) Determine the molarity:

1441.351 g / 1.122 g/mL = 1296.179 mL = 1.296179 L

4.500 mol / 1.296179 L = 3.472 M


Example #5: Given a density of 1.836 g/mL and a mass percent of H2SO4 of 96.00%, find the molarity, molality, and mole fraction. (This is the example #1 question, but done from a different starting point.)

Solution:

1) Assume 100.0 g of the solution is present. Determine the mass of each solution component:

H2SO4 ---> (100.0 g) (0.9600) = 96.00 g
H2O ---> (100.0 g) (0.0400) = 4.00 g

2) Determine the mole fraction:

H2SO4 ---> 96.00 g / 98.078 g/mol = 0.978813 mol
H2O ---> 4.00 g / 18.015 g/mol = 0.222037 mol

0.978813 mol + 0.222037 mol = 1.20085 mol

H2SO4 ---> 0.978813 mol / 1.20085 mol = 0.8151
H2O ---> 0.222037 mol / 1.20085 mol = 0.1849

or, the H2O can be obtained by subtraction:

1 - 0.8151 = 0.1849

3) Determine the molality:

4.00 g = 0.00400 kg

0.978813 mol / 0.00400 kg = 244.7 m

4) Determine the molarity:

100.0 g / 1.836 g/mL = 54.46623 mL = 0.05446623 L

0.978813 mol / 0.05446623 L = 17.97 M


Notice that the density is given in each of the above examples. This provides a necessary bridge between the volume-based concentration unit of molarity and the other concentration units (molality, mole fraction and mass percent) none of which uses volume in its definition.

If you were only given two of these: (1) mass percent or (2) molality or (3) mole fraction (and no density), you could not get the molarity (or the density).

However, suppose you are given one of these: (1) mass percent or (2) molality or (3) mole fraction and then also given molarity instead of the density. Could you get to the density and the other units? In other words, can we swap density and molarity in the given part of the problem? Let's see . . . .


Example #6: Reagent grade nitric acid is 70.40% HNO3 (63.0119 g/mol) by mass and its molarity is 16.00 M. Calculate the density, molality and mole fraction of nitric acid in the solution.

Solution:

1) 16.0 molar means this:

16.0 moles of HNO3
1.000 L (which equals 1000. mL) of solution.

2) The key point is that the 16.00 moles of HNO3 is 70.40% of the entire mass of the 1000. mL of solution.

(63.0119 g/mol) (16.0 mol) = 1008.19 g

1008.19 g / 0.7040 = 1432 g (this is the total mass of the solution)

3) The density is:

1432 g / 1000. mL = 1.432 g/mL

4) Molality:

1432 g - 1008.19 g = 423.81 g = 0.42381 kg

16.00 mol / 0.42381 kg = 37.75 m

5) Mole fraction

423.81 g / 18.015 g/mol = 23.5254 mol (of water)

16.00 mol + 23.5254 mol = 39.5254 mol (total moles)

23.5254 mol / 39.5254 mol = 0.5952 (mole fraction of water)

1 - 0.5952 = 0.4048 (mole fraction of the nitric acid)


Here's the question again:

. . . can we swap density and molarity in the given part of the problem?

And the answer is a very firm YES. Let's do another . . . .


Example #7: Reagent grade nitric acid (HNO3, MW = 63.0119 g/mol) has a molarity of 16.00 M and its molality is 37.75 m. Calculate the density, mass percent and mole fraction of nitric acid in the solution.

Solution:

1) Use the molality:

37.75 mol (of HNO3)
1.000 kg of solvent (water)

2) Compute the mass of the above solution:

(37.75 mol) (63.0119 g/mol) = 2378.7 g

1000 g + 2378.7 g = 3378.7 g

3) Compute the mass percent (I'll do just the nitric acid):

2378.7 g / 3378.7 g = 70.40%

4) Compute the mole fraction of the nitric acid:

HNO3 ---> 37.75 mol
H2O ---> 1000 g / 18.015 g/mol = 55.509 mol

55.509 mol + 37.75 mol = 93.259 mol

37.75 mol / 93.259 mol = 0.4048

5) Use the molarity to get the density of the solution:

37.75 mol / x = 16.00 mol/L

x = 2.359375 L

3378.7 g / 2359.375 mL = 1.432 g/mL


Turns out everything works just fine if the density and the molarity are the two values given. See Example #3 for another like #8.


Example #8: Reagent grade nitric acid (HNO3, MW = 63.0119 g/mol) has a molarity of 16.00 M and a density of 1.432 g/mL. Calculate the molality, mass percent and mole fraction of nitric acid in the solution.

Solution:

1) Assume 1.000 L of the solution is present. Determine its mass:

(1.432 g/mL) (1000. mL) = 1432 g

2) Determine the mass percent (just the nitric acid):

(16.00 mol) (63.0119 g/mol) = 1008.19 g

1008.19 g / 1432 g = 70.40%

3) Molality:

1432 - 1008.19 = 423.81 g = 0.42381 kg

16.00 mol / 0.42381 kg = 37.75 m

4) Mole ratio (of just the water):

HNO3 ---> 1008.19 g / 63.0119 g/mol = 16.00 mol
H2O ---> 423.81 g / 18.015 g = 23.5254 mol

23.5254 mol + 16.00 mol = 39.5254 mol

23.5254 / 39.5254 = 0.5952


Example #9: What is the molarity of a 30.0% (w/w) hydrogen peroxide solution?

Comment: note how the density has to be looked up in order to solve the problem. It's quite possible that this was deliberately done by the quesion writer (which was not the ChemTeam). If you do not pick that you have to supply the density, you'd think it's an impossible question.

Solution:

1) Looking on the Internet, the density is found to be 1.11 g/mL.

2) We will assume 1.00 L of the solution is present. (This is a convenient volume to take because you want molarity, which is defined as moles solute / liter solution.)

3) Compute the mass of 1.00 L of solution:

(1.11 g/mL) (1000 mL) = 1110 g

4) Compute the mass of H2O2 in the liter of solution

(1110 g) (0.300) = 333 g H2O2

5) Compute moles of H2O2 in the solution:

333 g / 34.0138 g/mol = 9.79 mol

6) Compute the molarity:

9.79 mol / 1.00 L = 9.79 M

Example #10: A 1.55 m solution of glucose (C6H12O6) is present. Determine the mole ratio of each solution component as well as the mass percent.

Solution:

1) 1.55 m means 1.55 mole of glucose dissolved in 1.00 kg of water. Determine the mass percents:

glucose ---> (1.55 mole) (180.1548 g/mol) = 279.23994 g
H2O ---> 1000 g

1000 g + 279.23994 g = 1279.23994 g

glucose ---> 100 - 71.87 = 21.83%
H2O ---> 1000 g / 1279.23994 g = 78.17%

2) Determine the mole ratios:

glucose ---> 1.55 mole
H2O ---> 1000 g / 18.015 g/mol = 55.509

1.55 + 55.509 = 57.059 mol glucose ---> 1 - 0.9728 = 0.0272
H2O ---> 55.509 / 57.059 = 0.9728

Comment: Give me the density and I can compute the molarity. Give me the molarity and I can compute the density. However, since neither one is present, the above problem is as far as we can go.


Problems 1 - 10

Problems 11 - 25

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