Chemical Formula Worksheet With Answers Page 6
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40/12.01 = C = 3.33; 6.73/1.01 = 6.66 H; 16/52.28 = 3.27 O. CH
O = 30.03 so n =
2
60.02/30.03 = 2 so molecular formula = 2 x (CH
O) =
2
C
H
O
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COMBUSTION ANALYSIS
Menthol, the substance we can smell in mentholated cough drops, is composed of
C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO
and
2
0.1159 g of H
O. What is the empirical formula for menthol? (x is a multiplication
2
operator).
1 mol C
1 mol CO
= 0.006430 mol C
2
0.2829 g of CO
X
= 0.006430 mol CO
X
2
2
1 mol CO
44.0g CO
2
2
2 mol H
1 mol H
O
2
= 0.01288 mol H
0.1159 g of H
O X
= 0.006439 mol H
0 X
2
2
1 mol H
0
18.0g H
O
2
2
Now we need to convert the moles to grams of these elements
12.0g C
0.006430 mol c x
= 0.07716g C
1 mol C
1.0g H
0.001288 mol H x
= 0.01288g H
1 mol H
Find the mass of Oxygen by subtracting the C and H from the total mass of the sample
Total= mass C + mass H + mass O
0.1005g= 0.07716 g C + 0.01288 g H + mass O
mass O = 0.01046g O
Convert to moles of O
1 mol O
0.01046g O x
= 0.0006538 mol O
16.0g O
Finally find the mole ratio by dividing by the smallest quantity
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