Question

page is designed to help students practice written oblems, and is meant to be princ print command and show all work in the spaces wided Round your answer to the significant digits and show work problems, and is meant to be printed out. Hit the • Round your answer to the correct number of 1. Determine the empirical formula of a compound that contains 36.5% sodium. 38.1% oxygen mpound that contains 36.5% sodium, 25.4% sulfur, and An organic compound has an empirical formula of CH and a molecular mass of 78 u. What is the molecular formula? 3. A hydrated compound has an analysis of 18.29% Ca. 32.37% CI, and 49.34% H.O. What is the formula?

Answer 1

1) First calculate moles of Na, S and O.

Moles of Na = % of Na / Atomic mass of Na = 36.5 / 23 = 1.59

Moles of S = % of S / Atomic mass of S =25.4 / 32 = 0.794

Moles of O = % of O / Atomic mass of O =38.1 / 16 =2.38

Ratio of number of moles of Na : S : O is

1.59 / 0.794 = 2 mol Na

0.794 / 0.794 = 1 mol S

2.38 / 0.794 = 3 mol O

Hence, empirical formula is Na₂SO₃

2) Empirical formula of compound is CH.

Empirical formula mass = 12+1= 13 u

We have, Molecular formula = r x empirical formula

Where r is the ratio of molar mass of substance to the empirical formula mass.

Therefore, r = 78 u / 13 u = 6

Hence, Molecular formula = r x empirical formula = 6 x CH = C₆H₆

ANSWER : Molecular formula = C₆H₆

3) Moles of Ca = % of Ca / Atomic weight of Ca = 18.29 /40 = 0.457

Moles Cl = % of Cl / Atomic weight of Cl = 32.37 / 35.45 = 0.913

Moles of H2O = % of H2O / Molecular weight of H₂O = 49.34 / 18 = 2.74

Ratio of number of moles of Ca : Cl : H₂O is

0.457 /0.457 = 1 mol Ca

0.913 / 0.457 = 2 mol Cl

2.74 / 0.457 = 6 mol H₂O

Hence, empirical formula = CaCl₂.6H₂O