A hydrocarbon, subjected to elemental analysis, was found to contain 79.89 % carbon and 20.11 % hydrogen by mass. What is the empirical formula of the hydrocarbon?

CH₃ is the empyrical formula for the hydrocarbon, a methyl group

Explanation:

Centesimal composition of hydrocarbon (X g / 100g of compound)

79.89 % C and 20.11 % H

We divide the mass of the composition, by the molar mass of the element:

79.89 g / 12 g/mol =  6.66 moles C

20.11 g / 1 g/mol = 20.11 moles H

We divide again, by the lowest value of each. In this case, 6.66

6.66 moles of C / 6.66 = 1 mol C

20.11 moles of H / 6.66 = 3 mol H

Molecular formula is CH₃

In this case, molecular formula matches the empyrical one.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

i think its 5049.

i hope this !