Question

2. a) what is the empirical formula of an ingredient in bufferin table that has the percent composition C1:4.25%, 0:56 93% and Mg: 28.83 % by mass

b) An analysis of sample of an organic compound shows that it contains 39.9 % C, 6.9% H, and 53,2% 1. calculate the empirical formula of the compound
2. the relative molecular mass is 60 what is the molecular formula of the compound?​

Answer 1

Step-by-step explanation:

a)Take percentages and divide by mole wt ( from periodic table) of the corresponding element

C 14.25 / 12 = 1.1875

O 56.93 / 15.99 = 3.56

Mg 28.83/24.3 = 1.186 Divide by the smallest number

C 1.1875/1.186 = 1

O 3.56 / 1.186 = 3

Mg 1.186 / 1.186 = 1

C O3 Mg commonly written as Mg CO3 ( magnesium carbonate)