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This test covers the mole, particle and mole conversions, molar mass with elements, molar mass with compounds, empirical formula, molecular formula, percent composition, and vocabulary.

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Chemistry - Chapter Eleven Practice Test

Question 1 of 18

1

Calculate the mass of 1.000 mole of CaCl(2).

Select one of the following:

  • 110.986 g/mol

  • 118.56 g/mol

  • 342.15 g/mol

  • 106.323 g/mol

Explanation

Question 2 of 18

1

Calculate grams in 3.0000 moles of CO(2).

Select one of the following:

  • 132.03 g

  • 102.01 g

  • 124.09 g

  • 137.15 g

Explanation

Question 3 of 18

1

Calculate number of moles in 32.0 g of CH(4).

Select one of the following:

  • 1.87 mol

  • 1.99 mol

  • 2.11 mol

  • 2.03 mol

Explanation

Question 4 of 18

1

Determine mass in grams of 40.0 moles of Na(2)CO(3).

Select one of the following:

  • 4,240 g

  • 5,565 g

  • 4,145 g

  • 4,056 g

Explanation

Question 5 of 18

1

Calculate the percent composition of KNO(3).

Select one of the following:

  • Molar mass = 151.1 g/mol
    Potassium: (39.10 / 101.1) x 100 = 38.67%
    Nitrogen: (14.01 / 101.1) x 100 = 13.86%
    Oxygen: (48.00 / 101.1) x 100 = 12.58%

  • Molar mass = 164.1 g/mol
    Potassium: (39.10 / 101.1) x 100 = 38.67%
    Nitrogen: (14.01 / 101.1) x 100 = 18.96%
    Oxygen: (48.00 / 101.1) x 100 = 47.48%

  • Molar mass = 101.1 g/mol
    Potassium: (39.10 / 101.1) x 100 = 38.67%
    Nitrogen: (14.01 / 101.1) x 100 = 13.86%
    Oxygen: (48.00 / 101.1) x 100 = 47.48%

Explanation

Question 6 of 18

1

Calculate percent composition for H(2)SO(4).

Select one of the following:

  • Molar mass = 98.07 g/mol
    Hydrogen: (2.016 / 98.07) x 100 = 2.06%
    Sulfur: (32.06 / 98.07) x 100 = 32.69%
    Oxygen: (64.00 / 98.07) x 100 = 65.26%

  • Molar mass = 98.07 g/mol
    Hydrogen: (2.016 / 98.07) x 100 = 3.09%
    Sulfur: (32.06 / 98.07) x 100 = 54.12%
    Oxygen: (64.00 / 98.07) x 100 = 65.39%

  • Molar mass = 98.07 g/mol
    Hydrogen: (2.016 / 98.07) x 100 = 9.05%
    Sulfur: (32.06 / 98.07) x 100 = 44.75%
    Oxygen: (64.00 / 98.07) x 100 = 14.96%

Explanation

Question 7 of 18

1

An oxide of chromium is found to have the following percent composition: 68.4% Cr and and 31.6% O. What is the compound's empirical formula?

Select one of the following:

  • CrO(4)

  • Cr(6)O(2)

  • Cr(2)O(3)

  • Cr(5)O

Explanation

Question 8 of 18

1

A sample with a molar mass of 34.00 g/mol is found to consist of 0.44g H and 6.92g O. Find its molecular formula.

Select one of the following:

  • HO

  • H(2)O

Explanation

Question 9 of 18

1

The percent composition of a compound was found to be 63.5% silver, 8.2% nitrogen, and 28.3% oxygen. Determine the compound's empirical formula.

Select one of the following:

  • Ag(3)NO

  • Ag(2)NO(3)

  • Ag(4)NO

  • AgNO(3)

Explanation

Question 10 of 18

1

If 4.04g of N combine with 11.46g O to produce a compound with a molar mass of 108.0 g/mol, what is the molecular formula of this compound?

Select one of the following:

  • N(3)O(6)

  • N(2)O(5)

  • N(1)O(8)

  • N(2)O(3)

Explanation

Question 11 of 18

1

Select from the dropdown list to complete the text.

( Percent composition, Molar mass, Molecular formula, Empirical formula ) is the percent by mass of each element in a compound.

Explanation

Question 12 of 18

1

Select from the dropdown list to complete the text.

( Percent Composition, Avogadro's number, Molecular Formula, Molar Mass ) is the number 6.02 * 10^23, which is the number of representative particles in a mole.

Explanation

Question 13 of 18

1

Select from the dropdown list to complete the text.

( Molar mass, Molecular Formula, Avogadro's Number, Percent Composition ) is the mass in grams of one mole of any pure substance.

Explanation

Question 14 of 18

1

Select from the dropdown list to complete the text.

( molar mass, mole, hydrate, empirical formula ): the SI base unit used to measure the amount of a substance. It is the number of representative particles, carbon atoms, in exactly 12 g of pure carbon-12.

Explanation

Question 15 of 18

1

Select from the dropdown list to complete the text.

( A hydrate, Molar mass, A mole, Empirical formula ) is a compound that has a specific number of water molecules bound to its atoms.

Explanation

Question 16 of 18

1

Select from the dropdown list to complete the text.

( Empirical formula, Molecular formula ) is a formula that specifies the actual number of atoms of each element in one molecule or formula unit of the substance.

Explanation

Question 17 of 18

1

Select from the dropdown lists to complete the text.

( Empirical formula, Molecular formula ) is a formula that shows the smallest whole number mole ratio of the elements of a compound and may or may not be the same as the ( empirical formula, molecular formula ).

Explanation

Question 18 of 18

1

Calculate percent composition of C(2)H(5)OH.

Select one of the following:

  • Molar mass = 46.07 g/mol
    Carbon: (24.022 / 46.07) x 100 = 12.64%
    Hydrogen: (6.048 / 46.07) x 100 = 13.13%
    Oxygen: (16.00 / 46.07) x 100 = 94.72%

  • Molar mass = 55.07 g/mol
    Carbon: (24.022 / 46.07) x 100 = 22.55%
    Hydrogen: (6.048 / 46.07) x 100 = 14.13%
    Oxygen: (16.00 / 46.07) x 100 = 34.73%

  • Molar mass = 86.07 g/mol
    Carbon: (24.022 / 46.07) x 100 = 52.14%
    Hydrogen: (6.048 / 46.07) x 100 = 13.13%
    Oxygen: (16.00 / 46.07) x 100 = 34.73%

Explanation

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