## Calculations involving density, percent concentration, molality and molarity

Example #1: Phosphoric acid is usually obtained as an 87.0% phosphoric acid solution. If it is 13.0 M, what is the density of this solution? What is its molality?

Solution for density:

1) Determine the moles of H3PO4 in 100.0 grams of 87.0% solution:

87.0 g of the 100.0 g is H3PO4

moles H3PO4 = 87.0 g / 97.9937 g/mol = 0.8878 mol

2) Calculate the volume of 13.0 M solution which contains 0.8878 mol of H3PO4:

13.0 mol/L = 0.8878 mol / x

x = 0.0683 L = 68.3 mL

3) Determine the density of the solution:

100.0 g / 68.3 mL = 1.464 g/mL = 1.46 g/mL (to three sig fig)

Solution for molality:

1) Let us assume 100.0 grams of solution. Therefore:

87.0 g is H3PO4
13.0 g is H2O

2) Calculate the molality:

moles H3PO4 = 87.0 g / 97.9937 g/mol = 0.8878 mol
kg of water = 0.0130 kg

molality = 0.8878 mol / 0.0130 kg = 68.3 molal

Example #2: Concentrated hydrochloric acid is usually available at a concentration of 37.7% by mass. What is its molar concentration? (The density of the solution is 1.19 g/mL.)

Solution Path #1:

1) Determine moles of HCl in 100.0 g of 37.7% solution:

37.7 g of this solution is HCl

37.7 g / 36.4609 g/mol = 1.03398435 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume

1.19 g/mL = 100.0 g / x

x = 84.0336 mL

3) Determine molarity:

1.03398435 mol / 0.0840336 L = 12.3 M

Solution Path #2:

1) Assume 1.00 L of solution. Use density to get mass:

1.19 g/mL = x / 1000 mL

x = 1190 g

2) (a) Use percent mass to get mass of HCl, then (b) convert to moles:

1190 g times 0.377 = 448.63 g

448.63 g / 36.4609 g/mol = 12.3 mol

3) Determine molarity:

12.3 mol / 1.00 L = 12.3 M

Example #3: I have a bottle of NH3. Its strength is 32.0% and its density is 0.89 g/mL. How do I figure out the molarity?

Solution:

1) Assume 100.0 g of solution is present. Calculate moles of ammonia present:

32.0 grams of NH3 are present.

32.0 g / 17.0307 g/mol = 1.879 mol

2) Determine volume of 100.0 g of solution:

100.0 g / 0.87 g/mL = 112.36 mL

3) Calculate molarity:

1.879 mol / 0.11236 L = 16.7 M

Example #4: An aqueous solution of hydrofluoric acid is 30.0% HF, by mass, and has a density of 1.101 g cm-3. What are the molality and molarity of HF in this solution?

Solution for molality:

1) Let us assume 100.0 grams of solution. Therefore:

30.0 g is HF
70.0 g is H2O

2) Calculate the molality:

moles HF = 30.0 g / 20.0059 g/mol = 1.49956 mol
kg of water = 0.0700 kg

molality = 1.49956 mol / 0.0700 kg = 21.4 molal

Solution for molarity:

1) Determine moles of HF in 100.0 g of 30.0% solution:

30.0 g of this solution is HF

30.0 g / 20.0059 g/mol = 1.49956 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume

1.101 g/mL = 100.0 g / x

x = 90.8265 mL

3) Determine molarity:

1.49956 mol / 0.0908265 L = 16.5 M

Example #5: Concentrated nitric acid is a solution that is 70.4% HNO3 by mass. The density of this acid is 1.42 g/mL. What is the molarity and the molality of the acid?

Solution for molarity:

1) Determine moles of HNO3 in 100.0 g of 70.4% solution:

70.4 g of this solution is HNO3

70.4 g / 63.0119 g/mol = 1.11725 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume

1.42 g/mL = 100.0 g / x

x = 70.422535 mL

3) Determine molarity:

1.11725 mol / 0.070422535 L = 15.86 M = 15.9 M (to three sf)

Solution for molality:

1) Let us assume 100.0 grams of solution. Therefore:

70.4 g is HNO3
29.6 g is H2O

2) Calculate the molality:

moles HNO3 = 70.4 g / 63.0119 g/mol = 1.11725 mol
kg of water = 0.0296 kg

molality = 1.11725 mol / 0.0296 kg = 37.7 molal

Special bonus part to Example #5: In the above solution, what is the mole fraction of HNO3?

Solution:

1) Let us assume 100.0 grams of solution. Therefore:

70.4 g is HNO3
29.6 g is H2O

2) Determine the moles of each substance:

HNO3 = 1.11725 mol
H2O = 29.6 g / 18.0152 g/mol = 1.643057 mol

3) Determine the mole fraction of HNO3:

1.11725 mol / (1.11725 mol + 1.643057 mol) = 0.405

Example #6: The density of toluene (C7H8) is 0.867 g/mL, and the density of thiophene (C4H4S) is 1.065 g/mL. A solution is made by dissolving 9.660 g of thiophene in 260.0 mL of toluene.

a) Calculate the molality of thiophene in the solution.
b) Assuming that the volumes of the solute and solvent are additive, determine the molarity of thiophene in the solution.

Solution to part a:

1) Determine the moles of thiophene:

9.660 g / 84.142 g/mol = 0.1148 mol

2) Determine mass of toluene in 260.0 mL

260.0 mL x 0.867 g/mL = 225.42 g

3) Calculate the molality:

0.1148 mol / 0.22542 kg = 0.509 molal

Solution to part b:

1) Determine volume of thiophene:

9.660 g ÷ 1.065 g/mL = 9.07 mL

2) Determine total volume:

260.0 mL + 9.07 mL = 269.07 mL

3) Calculate molarity:

0.1148 mol / 0.26907 L = 0.427 M

Example #7: An aqueous acetic acid solution is simultaneously 6.0835 molar and 8.9660 molal. Compute the density of this solution.

Solution:

1) Use molality to get moles:

8.9660 molal = x moles / 1.00 kg solvent

x = 8.9660 moles

2) Compute the mass of the above solution:

1000 g solvent + (8.9660 moles x 60.05 g/mol) = 1538.41 g

3) Use molarity to get the volume of solution containing 8.9660 moles

6.0835 mol/L = 8.9660 moles / x

x = 1.47382 L

4) Compute the density:
1538.41 g / 1473.82 mL = 1.0438 g/mL

Example #8: What is the density (in g/mL) of a 3.60 M aqueous sulfuric acid solution that is 29.0% H2SO4 by mass?

Solution:

1) Assume 100.0 g of solution is present.

2) Detemine the volume of solution that weighs 100.0 g:

29.0 g of the 100.0 g is H2SO4.

use MV = grams / molar mass to determine the volume.

(3.60 mol/L) (x) = 29.0 g / 98.1 g/mol

x = 0.0821 L

3) Determine the density:

100.0 g / 82.1 mL = 1.22 g/mL

Example #9: The density of an aqueous solution of nitric acid is 1.430 g/mL. If this solution contained 36.00% nitric acid by mass, how many mL of the solution would be needed to supply 150.20 grams of nitric acid?

Solution path #1:

1) 1.000 mL of solution weight 1.430 g, of which 36.00% is HNO3

2) 1.430 x 0.3600 tells you the grams of HNO3 in each mL of solution.

3) 150.20 g divided by the grams of HNO3 per one mL of solution.

Solution path #2:

1) Determine how many mL of the solution you need:

150.20 g nitric acid x (1 g solution/0.3600 g nitric acid) = 417.2 g solution.

2) Determine the volume of solution that weighs 417.2 g:

417.2 g solution x (1 mL/1.430 g solution)

Example #10: A bottle of commercial sulfuric acid (density 1.787 g/cc) is labeled as 86% by weight. What is the molarity of acid?

Solution:

1) Determine moles of H2SO4 in 100.0 g of 86% solution:

86 g of this solution is H2SO4

86 g / 98.08 g/mol = 0.876835237 mol

2) Determine volume of 100.0 g of solution:

density = mass / volume

1.787 g/mL = 100.0 g / x

x = 55.96 mL

3) Determine molarity:

0.876835237 mol / 0.05596 L = 15.67 M = 15.7 M (to three sf)

Note: Try this problem with 96.0% and a density of 1.84 g/mL. The answer is 18.0 M.

Example #11: Calculate the molarity and mole fraction of acetone in a 2.28-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density of acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume that the volumes of acetone and ethanol add.

Solution for molarity:

Remember, 2.28-molal means 2.28 moles of acetone in 1.00 kilogram of ethanol.

1) Determine volumes of acetone and ethanol, then total volume:

acetone

2.28 mol x 58.0794 g/mol = 132.421 g

132.421 g divided by 0.788 g/cm3 = 168.047 cm3

ethanol

1000 g divided by 0.789 g/cm3 = 1267.427 cm3

total volume

168.047 + 1267.427 = 1435.474 cm3

2) Determine molarity:

2.28 mol / 1.435 L = 1.59 M

Solution for mole fraction:

1) Determine moles of ethanol:

1000 g / 46.0684 g/mol = 21.71 mol

2) Determine mole fraction of acetone:

2.28 / (2.28 + 21.71) = 0.0950

Example #12: Calculate the normality of a 4.0 molal sulfuric acid solution with a density of 1.2 g/mL.

Reminders:

N = #equivalents / L solution
#equivalents = molecular weight / n (n = number of H+ or OH¯ released per dissociation.)
molal = moles solute / kg solvent

Solution:

1) Determine grams of H2SO4 present:

4.0 molal = 4.0 moles H2SO4 / 1000 g solution

4.0 mol times 98.09 g/mol = 392.32 g

2) Determine equivalent weight for H2SO4:

98.09 g/mol / 2 dissociable hydrogen/mol = 49.05 g/equivalent

3) Determine # equivalents in 392.32 g:

392.32 g times (1 equivalent / 49.05 g) = 8.0 equivalents

4) Determine volume of solution:

392.32 g + 1000 g = 1392.32 g (total mass of solution)

1392.32 g / 1.2 g/mL = 1160.27 mL

5) Determine normality:

N = 8.0 equivalents / 1.16027 L = 6.9 N

Example #13: An car antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL, molar mass 62.07 g/mol) and water (d = 1.000 g/mL) at 20.0 °C. The density of the solution is 1.070 g/mL.

Express the concentration of ethylene glycol as:

(a) volume percent
(b) mass percent
(c) molarity
(d) molality
(e) mole fraction

Solution to (a):

Since the volumes are equal, the volume percent of ethylene glycol is 50%

Solution to (b):

1) Determine the masses of equal volumes (we'll use 50.0 mL) of the two substances:

ethylene glycol: (50.0 mL) (1.114 g/mL) = 55.7 g
water: (50.0 mL) (1.000 g/mL) = 50.0 g

2) Determine percent due to ethylene glycol:

55.7 g / 105.7 g = 52.7%

Solution to (c):

1) Determine moles of ethylene glycol:

55.7 g / 62.07 g/mol = 0.89737 mol

2) Determine volume of solution:

105.7 g / 1.070 g/mL = 98.785 mL

3) Determine molarity:

0.89737 mol / 0.098785 L = 9.08 M

Solution to (d):

0.89737 mol / 0.050 kg = 17.9 m

Note the large difference between the molarity and the molality.

Solution to (e):

1) Determine moles of water:

50.0 g / 18.0 g/mol = 2.77778

2) Determine mole fraction for ethylene glycol:

0.89737 mol / 3.67515 mol = 0.244

Example #14: What is the percent of CsCl by mass in a 0.0711 M CsCl solution that has a density of 1.09 g/mL?

Solution:

1) Determine mass of dissolved CsCl:

Let us assume 100.0 mL of solution.

MV = grams / molar mass

(0.0711 mol/L) (0.100 L) = x / 168.363 g/mol

x = 1.197 g

2) Determine mass of solution:

1.09 g/mL times 100.0 mL = 109 g

Determine mass percent of CsCl in solution:

1.197 g / 109 g = 1.098%

to three sig figs: 1.10%