General Chemistry 10th Edition

Published by Cengage Learning
ISBN 10: 1-28505-137-8
ISBN 13: 978-1-28505-137-6

Chapter 3 - Calculations with Chemical Formulas and Equations - Questions and Problems - Page 121: 3.69

Answer

The empirical formula is $C_{3}H_{4}O_{2}$.

Work Step by Step

Strategy: 1- Use percentage composition to find the mass of each element in the compound. For this assume you have 100 g compound. If for example you are given 21% as the percentage composition of an element X in the compound, this is $\frac{21}{100}\times100g$ = 21 g of element X since you assumed a mass 100 g for the compound. Thus the mass of the element equals the numerical value of percentage composition. 2- Convert the masses to moles, using conversion factor 3- Divide each mole number by the smallest one in order to find the smallest integers. 4- If you do not get integers, than find a whole number to multiply the results of step 3 to get integers. The compound contains these percentages by mass of each element: 50.0% C, 5.6% H, and 44.4% O. Assuming 100 g compound we get the masses of each element: 50.0 g C, 5.6 g H, 44.4 g O. Now we convert masses to moles: Moles of C: 50.0 g$\times\frac{1 mol(C)}{12.0 g (C)}$ = 4.17 mol C Moles of H: 5.6 g$\times\frac{1 mol(H)}{1.00 g (H)}$ = 5.60 mol H Moles of oxygen: 44.4g$\times\frac{1 mol (O)}{16.0 g (O) }$ = 2.78 mol O Divide the mole number by the smallest one. For C: $\frac{4.17mol}{2.78mol}$ = 1.50 For H: $\frac{5.60mol}{2.78mol}$ = 2.02 For O: $\frac{2.78mol}{2.78mol}$ = 1.00 We round the last digit, that is subject to experimental error, and we get $C_{1.5}H_{2}O_{1}$. We see that subscript for C is not integers, but that one can be turned into integer by multiplying by 2. Remember we have to multiply by 2 the subscript for H and O as well in order to get ratio of elements in the compound. So we have for C: 1.5x2 = 3 for H : 2 x 2 = 4 for O: 1 x 2 = 2 So the empirical formula is $C_{3}H_{4}O_{2}$.
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