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Chemistry - Chapter Eleven Practice Test

Description

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.
kiaraisastudent
Quiz by kiaraisastudent, updated more than 1 year ago
kiaraisastudent
Created by kiaraisastudent about 8 years ago
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Resource summary

Question 1

Question
Calculate the mass of 1.000 mole of CaCl(2).
Answer
  • 110.986 g/mol
  • 118.56 g/mol
  • 342.15 g/mol
  • 106.323 g/mol

Question 2

Question
Calculate grams in 3.0000 moles of CO(2).
Answer
  • 132.03 g
  • 102.01 g
  • 124.09 g
  • 137.15 g

Question 3

Question
Calculate number of moles in 32.0 g of CH(4).
Answer
  • 1.87 mol
  • 1.99 mol
  • 2.11 mol
  • 2.03 mol

Question 4

Question
Determine mass in grams of 40.0 moles of Na(2)CO(3).
Answer
  • 4,240 g
  • 5,565 g
  • 4,145 g
  • 4,056 g

Question 5

Question
Calculate the percent composition of KNO(3).
Answer
  • 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%

Question 6

Question
Calculate percent composition for H(2)SO(4).
Answer
  • 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%

Question 7

Question
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?
Answer
  • CrO(4)
  • Cr(6)O(2)
  • Cr(2)O(3)
  • Cr(5)O

Question 8

Question
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.
Answer
  • HO
  • H(2)O

Question 9

Question
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.
Answer
  • Ag(3)NO
  • Ag(2)NO(3)
  • Ag(4)NO
  • AgNO(3)

Question 10

Question
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?
Answer
  • N(3)O(6)
  • N(2)O(5)
  • N(1)O(8)
  • N(2)O(3)

Question 11

Question
[blank_start]Percent composition[blank_end] is the percent by mass of each element in a compound.
Answer
  • Percent composition
  • Molar mass
  • Molecular formula
  • Empirical formula

Question 12

Question
[blank_start]Avogadro's number[blank_end] is the number 6.02 * 10^23, which is the number of representative particles in a mole.
Answer
  • Percent Composition
  • Avogadro's number
  • Molecular Formula
  • Molar Mass

Question 13

Question
[blank_start]Molar mass[blank_end] is the mass in grams of one mole of any pure substance.
Answer
  • Molar mass
  • Molecular Formula
  • Avogadro's Number
  • Percent Composition

Question 14

Question
[blank_start]mole[blank_end]: 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.
Answer
  • molar mass
  • mole
  • hydrate
  • empirical formula

Question 15

Question
[blank_start]A hydrate[blank_end] is a compound that has a specific number of water molecules bound to its atoms.
Answer
  • A hydrate
  • Molar mass
  • A mole
  • Empirical formula

Question 16

Question
[blank_start]Molecular formula[blank_end] is a formula that specifies the actual number of atoms of each element in one molecule or formula unit of the substance.
Answer
  • Empirical formula
  • Molecular formula

Question 17

Question
[blank_start]Empirical formula[blank_end] 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 [blank_start]molecular formula[blank_end].
Answer
  • Empirical formula
  • Molecular formula
  • empirical formula
  • molecular formula

Question 18

Question
Calculate percent composition of C(2)H(5)OH.
Answer
  • 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%
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