A compound is 37.48% C, 12.58% H, and 49.93% O by mass. Part A What is the empirical formula of the compound

CH4O

Explanation:

Empirical formula of a compound represents the smallest whole number ratio of its constituent elements.

To find the empirical formula, we say:

37.48% C = 37.48g of Carbon

12.58% H = 12.58g of Hydrogen

49.93% O = 49.93g of Oxygen

We convert each mass to mole by dividing by their respective atomic mass.

C=12g/mol, H=1g/mol, O=16g/mol

C = 37.48/12 = 3.123mol

H = 12.58/1= 12.58mol

O = 49.93/16 = 3.121mol

Next, we divide each mole value by the smallest (3.121mol)

C = 3.123/3.121 = 1.0006

H = 12.58/3.121 = 4.03

O = 3.121/3.121 = 1

Approximately, the empirical formula of

CHO is in ratio 1:4:1

Hence, CH4O is the empirical formula

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation: