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Chemistry Help?
Combustion analysis of 63.8 mg of a C, H and O containing compound produced 145.0 mg of CO2 and 59.38 mg of H2O. What is the empirical formula for the compound?
A) C3H7O B) C5H2O C) C3H6O D) C6HO3 E) CHO
1 Answer
- CarboraneLv 61 decade agoFavourite answer
( ) is the % mass of an element
Carbon: 145 mg x (12.01 g / 44.01 g) = 39.6 mg
Hydrogen: 59.38 mg x (2 x 1.008 g / 18.02 g) = 6.64 mg
why did we calculate these elements? because all carbon and hydrogen end up in these two compounds.
Mass of oxygen = 63.8 mg - 39.6 mg - 6.64 mg = 17.6 mg
With all masses for 3 elements you can calculate the empirical formula:
C: 39.6 mg * 1 mmol / 12.01 mg = 3.25 mmol
H: 6.64 mg * 1 mmol / 1.008 mg = 6.59 mmol
O: 17.6 mg * 1 mmol / 16.00 mg = 1.10 mmol
divide all three by the smallest number, 1.10
C:H:O = 2.95:5.99:1
The choice is C. C3 : H6 : O1
