A compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? the molecular mass of the compound is 132 amu. what is the molecular formula?

Question

Zion Graham

Chemistry

26 January, 16:49

A compound is 54.53% c, 9.15% h, and 36.32% o by mass. what is its empirical formula? the molecular mass of the compound is 132 amu. what is the molecular formula?

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Answers (1)

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Catalina Oneill · Answer 1

First - you need the empirical formula.

So, assume you have 100 g of the compound.

If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.

54.53 g C (1 mole C / 12.01 g C) = 4.540

9.15 g H (1 mole H / 1.008 g H) = 9.077

36.32 g O (1 mole O / 15.9994 g O) = 2.270

Take the smallest number found and divide the others by it to get the empirical formula.

4.540/2.270 = 2.

9.077/2.270 = 4.

2.270/2.270 = 1.

So, that gives you the empirical formula of C2H4O.

Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.

132/44 = 3.

So, 3 (C2 H4 O) = C6H12O3 = molecular formula.