Establish the chemical formula of a hydrocarbon with a relative molecular weight of 58.

Establish the chemical formula of a hydrocarbon with a relative molecular weight of 58. The mass fraction of carbohydrate in it is 82.72 percent, hydrogen is 17.44 percent, name the substance.

Given:
CxHy
Mr (CxHy) = 58
ω (C in CxHy) = 82.72%
ω (H in CxHy) = 17.44%

To find:
CxHy -?

Decision:
1) Calculate the number of carbon atoms in the substance:
N (C in CxHy) = (ω (C in CxHy) * Mr (CxHy)) / (Ar (C) * 100%) = (82.72% * 58) / (12 * 100%) = 4;
2) Calculate the number of hydrogen atoms in the substance:
N (H in CxHy) = (ω (H in CxHy) * Mr (CxHy)) / (Ar (H) * 100%) = (17.44% * 58) / (1 * 100%) = 10;
3) Establish the formula for an unknown substance:
Unknown substance – C4H10 – butane.

Answer: Unknown substance – C4H10 – butane.