Acompound is composed of only c, h, and o. the combustion of a .519-g sample of the compound yields 1.24 g of co_2 and .255 g of h_2o. what is the empirical formula of the compound?

Empirical formula is C₃H₃O.

Explanation:

Given data:

Mass of compound = 0.519 g

Mass of CO₂ = 1.24 g

Mass of H₂O = 0.255 g

Empirical formula = ?

Solution:

%age of C,H,O

C = 1.24 g/0.519 × 12/44 ×100 = 65.5%

H = 0.255 g/0.519 × 2.016/18 ×100 =  5.6%

O = 100 - (65.5+5.6)

O = 28.9%

Number of gram atoms of H = 5.6 / 1.01 = 5.5

Number of gram atoms of O =  28.9 / 16 = 1.81

Number of gram atoms of C = 65.5 / 12 = 5.5

Atomic ratio:

            C                      :        H              :         O

          5.5/1.81              :       5.5/1.81     :       1.81/1.81

            3                      :        3              :        1

C : H : O = 3 : 3 : 1

Empirical formula is C₃H₃O.

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

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