Take 100 g sample which gives
40 g C
6.71 g H
53.29 g O
Convert to mols by mols = grams/atomic mass
40/12 = approx 3.3
6.71/1 = 6.71
53.29/16 = approx 3.3
To determine the empirical formula, find the ratio of these elements to each other with the smallest number being 1.00. The easy way to do that is to divide the smallest number by itself (thereby assuring 1.00 for that one) and divide all of the other numbers by the same small number. So
3.3/3.3 = 1.00 C
6.71/3.3 = 2 H
3.3/3/3 = 1 O
emprical formula is CH2O
From the 0.167 g sample contg 0.00278 mols we an use mol = g/molar mass or
molar mass = g/mol = 0.167/0.00278 = 60.07 for the molar mass.
Then empirical formula mass is 12+2+16 = 30
empirical mass x whole number = molar mass
30 x whole number = 60
whole number = 60/30 = 2
So the molecular formula is (CH2O)2 or C2H4O2 and the molar mass is
2*12.01 + 2*1 + 1*16 = ?
A 0.167-g sample of an unknown compound contains 0.00278 mol of the compound. Elemental analysis of the acid gives the following percentages by mass: 40.00% C; 6.71% H; 53.29% O.
1.)Determine the molecular formula of this acid.
2.)Determine the molar mass of the unknown acid.
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