What is the empirical formula of a compound that contains only Carbon, Hydrogen, and Oxygen with the following elemental analysis: C (40.0%), H (6.67%) ?
How do I solve this? Thank you!!!
How do I solve this? Thank you!!!
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Take a hypothetical sample of 100.0 grams:
(40.0 g C) / (12.01078 g C/mol) = 3.3303 mol C
(6.67 g H) / (1.007947 g H/mol) = 6.6174 mol H
100g - 40.0g - 6.67g = 53.33 g O
(53.33 g O) / (15.99943 g O/mol) = 3.3332 mol O
Divide by the smallest number of moles:
(3.3303 mol C) / 3.3303 = 1.00
(6.6174 mol H) / 3.3303 = 1.99
(3.3332 mol O) / 3.3303 = 1.00
Round to the nearest whole numbers to find the empirical formula:
CH2O
(40.0 g C) / (12.01078 g C/mol) = 3.3303 mol C
(6.67 g H) / (1.007947 g H/mol) = 6.6174 mol H
100g - 40.0g - 6.67g = 53.33 g O
(53.33 g O) / (15.99943 g O/mol) = 3.3332 mol O
Divide by the smallest number of moles:
(3.3303 mol C) / 3.3303 = 1.00
(6.6174 mol H) / 3.3303 = 1.99
(3.3332 mol O) / 3.3303 = 1.00
Round to the nearest whole numbers to find the empirical formula:
CH2O