1.15 Molarity  (Page 2/14)

Calculating molar concentrations from the mass of solute

Distilled white vinegar ( [link] ) is a solution of acetic acid, CH ₃ CO ₂ H, in water. A 0.500-L vinegar solution contains 25.2 g of acetic acid. What is the concentration of the acetic acid solution in units of molarity?

Solution

As in previous examples, the definition of molarity is the primary equation used to calculate the quantity sought. In this case, the mass of solute is provided instead of its molar amount, so we must use the solute’s molar mass to obtain the amount of solute in moles:

$M=\frac{\text{mol solute}}{\text{L solution}}=\frac{\mathrm{25.2\; g}{\text{CH}}_3{\text{CO}}_2\text{H}\times \frac{1{\text{mol CH}}_2{\text{CO}}_2\text{H}}{{\text{60.052 g CH}}_2{\text{CO}}_2\text{H}}}{\text{0.500 L solution}}=0.839M$

$\begin{array}{l}\\ M=\frac{\text{mol solute}}{\text{L solution}}=0.839M\\ M=\frac{0.839\text{mol solute}}{1.00\text{L solution}}\end{array}$

Check your learning

Calculate the molarity of 6.52 g of CoCl ₂ (128.9 g/mol) dissolved in an aqueous solution with a total volume of 75.0 mL.

Answer:

0.674 M

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When performing calculations stepwise, as in [link] , it is important to refrain from rounding any intermediate calculation results, which can lead to rounding errors in the final result. In [link] , the molar amount of NaCl computed in the first step, 1.325 mol, would be properly rounded to 1.32 mol if it were to be reported; however, although the last digit (5) is not significant, it must be retained as a guard digit in the intermediate calculation. If we had not retained this guard digit, the final calculation for the mass of NaCl would have been 77.1 g, a difference of 0.3 g.

In addition to retaining a guard digit for intermediate calculations, we can also avoid rounding errors by performing computations in a single step (see [link] ). This eliminates intermediate steps so that only the final result is rounded.