1. A 3.285-g sample of an oxide of Ru contains 2.655 g Ru. What is the empirical

Question

1. A 3.285-g sample of an oxide of Ru contains 2.655 g Ru. What is the empirical formula of the oxide?

2. What is the empirical formula of a hydrocarbon if complete combustion or 1.700 mg of the hydrocarbon produced 5.602 mg of CO2 and 1.529 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 80

molecular formula =

3. What is the atomic mass of element X if 5.81 g XCl3 contains 2.07 g X?

atomic mass of X =

4. Capsaicin is C18H27N1O3. What mass of Capsaicin contains 0.626 moles of C? ______ g

Explanation / Answer

1) Determine mass:

Ru 2.655 g
O 3.285 g - 2.655 g = 0.63 g

Determine moles:

Ru 2.655 g / 101.07 g/mol = 0.02626 mol
O 0.63 g / 16.00 g/mol = 0.039375 mol

Ru : O = 1 : 1.5

Hence Ru2O3 is the empirical formula

2) All the carbon from the hydrocarbon ends up in the CO2, so you can determine moles of C in the 1.700 mg of hydrocarbon from the mass of CO2 produced.

moles CO2 = mass / molar mass = 5.602 mg / 44.01 g/mol = 0.1273 mmol
each CO2 has 1 C, therefore moles C = moles CO2 = 0.1273 mmol

All the H ends up in the H2O
moles H2O = 1.529 mg / 18.00 g/mol = 0.08494 mmol
each H2O has 2 H, therefore moles H = 2 x moles H2O = 2 x 0.08494 mmol = 0.1698 mmol

the ratio of mmol C : mmol H
= 0.1273 : 0.1698

divide each number in the ratio by the smallest number
0.1273 / 0.1273 : 0.1698 / 0.1273
= 1 : 1.33

This is not a whole number ratio, multiply it by the smallest number needed to make each number a whole number. In this case x 3

3 x (1 : 1.33)
= 3 : 4

empirical formula
C3H4

To find the molecular formula divide the molecular mass by the formula mass of the empirical formula. This tells you how many times the empirical formula fits into the molecular formula.

80 g/mol / 40.00 = 2
The molecular formula is 4 x the empirical formula

2 x C3H4

molecular formula
C6H8

3) XCl3 = 5.81 g , X = 2.07 g

3 Cl = 5.81 g - 2.07 g = 3.74 g,

Cl = 3.74 / 3 = 1.25 g

Moles of Cl = 1.25 / 35.45 g/mol = 0.03516 mol

Therefore [3 Cl] = [X] => 0.03516 mol = 2.07 g X

atomic mass of X = 2.07 g / 0.03516 mol = 58.86 g/mol ( It must be CoCl3)

4) Molar mass of C18H27N1O3 = (18 x 12) + (27 x 1) + (1 x 14) + (3 x 16) = 305 g/mol

Mass = 305 g/mol X 0.626 mol = 190.93 g

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