A compound is composed of the elements, carbon, hydrogen, and fluorine only. Combustion of the compound produces CO2, H2O, and HF C?H?F? + O2 ➝ CO2 + H2O + HF Combustion of a 5.000 g sample of the compound produces the products shown in the table below. Product Mass (grams) CO2 8.724 H2O 2.143 HF 2.379 Determine the empirical formula for the compound.
A compound is composed of the elements, carbon, hydrogen, and fluorine only. Combustion of the compound produces CO2, H2O, and HF
C?H?F? + O2 ➝ CO2 + H2O + HF
Combustion of a 5.000 g sample of the compound produces the products shown in the table below.
| Product | Mass (grams) |
|---|---|
| CO2 | 8.724 |
| H2O | 2.143 |
| HF | 2.379 |
Determine the empirical formula for the compound.
Assuming the empirical formula of compound to be CaHbFc.
Given : mass of CO2 produced = 15.182 g
Mass of H2O produced = 3.625 g
Mass of HF produced = 3.451 g
Molar mass of CO2 = Atomic mass of C + Atomic mass of O X 2 = 12 + 16 X 2 = 44 g/mol
Since mass = moles X molar mass
=> 15.182 = moles of CO2 X 44
=> moles of CO2 = 0.345 mol
Since each molecule of CO2 is having 1 atom of C.
Hence moles of C in CO2 = moles of CO2 = 0.345 mol
Since in the combustion reaction, all the C in CO2 is coming from CaHbFc.
Hence moles of C in CaHbFc = moles of C in CO2 = 0.345 mol
Molar mass of HF = Atomic mass of H + Atomic mass of F = 1 + 19 = 20 g/mol
Since mass = moles X molar mass
=> 3.451 = moles of HF X 20
=> moles of HF = 0.17255 mol
Since each molecule of HF is having 1 atom of F and 1 atom of H.
Hence moles of F in HF = moles of HF = 0.17255 mol
And moles of H in HF = moles of HF = 0.17255 mol
Since in the combustion reaction, all the F in HF is coming from CaHbFc.
Hence moles of F in CaHbFc = moles of F in HF = 0.17255 mol
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