Question

12) An organic compound contains C, H, and O. The percentages of H and C are 6.73% H, and 39.99% C by mass. The molar mass is 60.06 u. What is the molecular formula of the compound? A) CH2O B) C3H6O3 C) C2H3O2 D) C2H4O2 E) C4H8O4

1) A 10.0 g sample of bismuth tribromide, BiBr3, contains: A) 5.360 × 1022 total number of ions B) 0.322 mol BiBr3 C) 3.14 × 1022 formula units BiBr3 D) 1.34 × 1022 bromide ions E) 4.020 × 1022 total number of ions

can you show me how to solve these questions???

Answer 1

12) In the organic compound

Percentage of H = 6.37 %; Percentage of C = 39.99 %; Hence percentage of O = 100 - (6.37 + 39.99) = 53.64 %

Element	Percentage	Atomic mass	Relative no of atoms	simplest ratio
C	39.99	12	39.99/12 = 3.33	3.33/3.33 = 1
H	6.37	1	6.37/1 = 6.37	6.37 / 3.33 2
O	53.64	16	53.64/16 = 3.35	3.35/3.33 1

Hence, the empirical formula = CH₂O

Empirical formula mass = 30 u

n= Molecular mass / Empirical formula mass = 60.06 / 30 2

Hence, molecular formula = (CH₂O)₂ = C₂H₄O₂ = CH₃COOH