# Chapter 3 - Chemical Reactions - Questions for Review and Thought - General Questions - Page 148h: 109

Percent impurity = 6.26%

#### Work Step by Step

1000ml = 1L 21.3ml = 0.0213 L 200ml = 0.200 L 1. Find the volume necessary for an equal number of moles: $C_1 * V_1 = V_2 * C_2$ $0.125* 0.0213= 0.200 * C_2$ $0.00266 = 0.200 * C_2$ $C_2 = 0.01331$ 2. Find the number of moles: $Concentration(M) = \frac{n(mol)}{V(L)}$ $0.01331 = \frac{n(mol)}{0.200}$ $0.01331 * 0.200 = n(mol)$ $2.662 \times 10^{-3} moles = n(mol)$ 3. Determine the molar mass of this compound $(HC_6H_7O_6)$: 1.01* 1 + 12.01* 6 + 1.01* 7 + 16* 6 = 176.08g/mol 4. Calculate the mass $mm(g/mol) = \frac{mass(g)}{n(mol)}$ $mm(g/mol) * n(mol) = mass(g)$ $176.08 * 2.662 \times 10^{-3} = mass(g)$ $0.4687 = mass(g)$ 5. Now, calculate the percent of the mass that reacted: $\frac{0.4687g}{0.5g} \times 100\% = 93.74\%$ This is the purity; we have to find the impurity: 100% - 93.74% = 6.26%

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