Question

1.

a. The compound X₃Y is 35.0% X. What is the molar mass of Y if the molar mass of X is 62.4 g/mol?

b. What is the atomic mass of element X if 9.94 g XCl₃ contains 3.26 g X?

c. A 2.953-g sample of an oxide of V contains 1.654 g V. What is the empirical formula of the oxide?

d. What is the empirical formula of a hydrocarbon if complete combustion or 6.900 mg of the hydrocarbon produced 22.738 mg of CO₂ and 6.205 mg of H₂O? Be sure to write C first in the formula.

empirical formula =

What is the molecular formula if the molar mass of the hydrocarbon is found to be about 120

molecular formula =

Answer 1

Answer – a) Given, compound X₃Y , percent of X = 35.0 % , molar mass of x = 62.4 g/mol , molar mass of Y = ?

In the compound X₃Y there are 3 X and X molar mass = 62.4 g/mol

So in the compound X₃Y total mass of X = 3 * 62.4 g = 187.2 g

Now we have % of X = 35 % and we know

% of X = mass of X / total mass* 100 %

So, total mass = mass of X / % of X * 100 %

                        = 187.2 g / 35 % *100 %

                         = 534.85 g

So, mass of Y = total mass – mass of X

                        = 534.85 – 187.2

                        = 347.6 g

So, molar mass of Y = 347.6 g/mol.

b) Given, mass of XCl₃ = 9.94 g , mass of X = 3.26 g

so, mass of Cl in the compound = 9.94 g – 3.26 g

                                                        = 6.68 g

Moles of Cl = 6.68 g / 35.5 g.mol^-1

                     = 0.188 mole

In the XCl₃

3 moles of Cl = 1 moles of X

So, 0.188 moles of Cl = ?

= 0.0627 moles of X

So, atomic mass X = 3.26 g / 0.0627 mol

                               = 51.97 g/mol

c) Given, mass of oxide of V = 2.953 g , mass of V = 1.654 g

First we need to calculate the mass of O in the oxide of the V

Mass of O = mass of oxide – mass of V

                   = 2.953 – 1.654

                    = 1.299 g

Moles of V = 1.654 g / 50.942 g.mol^-1

                   = 0.0325 moles

Moles of O = 1.299 g / 15.998 g.mol^-1

                   = 0.0812 moles

So, V = 0.0325 / 0.0325 = 1

O = 0.0812 / 0.0325 = 2.5

So we need to multiply by with 2 for converting the fraction number of O to whole number.

V = 1*2 = 2

O = 2.5*2 = 5

So empirical formula is V₂O₅

d) Given, mass of compound = 6.900 mg = 0.006900 mg

Mass of CO₂ = 22.738 mg = 0.022738 g ,

mas of H₂O = 6.205 mg = 0.006205 g

First we need to calculate moles of CO₂ and H₂O from the given mass

Moles of CO₂ = 0.022738 g / 44.0 g.mol^-1 = 0.000517 mole

Moles of H₂O = 0.006205 g / 18.015 g.mol^-1 = 0.000344 moles

Moles of C from the moles of CO₂

1 moles of CO₂ = 1 moles of C

So, 0.000517 moles of CO₂ = ?

= 0.000517 moles of C

Moles of H from moles of H₂O

1 moles of H₂O = 2 moles of H

So, 0.000344 moles of H₂O = ?

= 0.000689 moles of H

So moles of C and O both are smallest, so we need to divided each mole by this mole

So , C = 0.000517 / 0.000517 = 1

      H = 0.000689 / 0.000517 = 1.33

Now we multiply with 3 for converting the fraction number to whole number

C = 1*3 = 3

H = 1.33*3 = 4.0

So empirical formula is C₃H₄

Molecular formula = n * empirical formula

n = molecular formula mass / empirical formula mass

= 120 / 40

= 3

So molecular formula = 3 * C₃H₄

                                   = C₉H₁₂