An anylysis of an organic compound found in meteriotes shows the following percentage composition in mass
C = 37.5
H = 12.5
O = 50
Find the empirical formula???
Chemistry calculation help?
2016-05-16 6:36 pm
回答 (3)
2016-05-16 6:49 pm
✔ 最佳答案
Percentage = amount in grams in 100 g.37.5 g / 12.011 g C mol-1 = 3.122 moles C.
12.5 g / 1.00 g H mol-1 = 12.5 moles H.
50 g / 15.99 g O mol-1 = 3.127 moles O.
Divide by lowest: 3.122 so molar ratio:
1 mol C, 4 mol H, 1 mol O.
Empirical formula: CH4O.
2016-05-16 6:48 pm
Atomic weights: C=12, H=1, O=16
37.5/12 = 3.125
12.5/1 = 12.5
50/16 = 3.125
Divide through by the smallest:
3.125/3.125 = 1
12.5/3.125 = 4
3.125/3.125 = 1
Empirical formula: CH4O
37.5/12 = 3.125
12.5/1 = 12.5
50/16 = 3.125
Divide through by the smallest:
3.125/3.125 = 1
12.5/3.125 = 4
3.125/3.125 = 1
Empirical formula: CH4O
2016-05-16 6:47 pm
Molar mass of C = 12.0 g/mol
Molar mass of H = 1.0 g/mol
Molar mass of O = 16.0 g/mol
Mole ratio C : H : O
= 37.5/12.0 : 12.5/1.0 : 50/16.0
= 3.13 : 12.5 : 3.13
= 1 : 4 : 1
Empirical formula = CH₄O
Molar mass of H = 1.0 g/mol
Molar mass of O = 16.0 g/mol
Mole ratio C : H : O
= 37.5/12.0 : 12.5/1.0 : 50/16.0
= 3.13 : 12.5 : 3.13
= 1 : 4 : 1
Empirical formula = CH₄O
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