Module 05: Stoichiometry Flashcards

Question 1

Q

01.05 The Mole Concept

What is true about 1.0 mol Ca and 1.0 mol Mg? (3 points)

They are equal in mass.
They contain the same number of atoms.
They have the same atomic mass.
Their molar masses are equal.

Answer

A

2. They contain the same number of atoms.

Question 2

Q

01.05 The Mole Concept

What is the molar mass of silver (Ag)? (3 points)

47.0 g/mol
60.02 g/mol
107.87 g/mol
196.97 g/mol

Answer

A

3. 107.87 g/mol

Question 3

Q

01.05 The Mole Concept

How many moles of copper (Cu) are in 65.8 g Cu? (3 points)

1.04 mol Cu
10.9 mol Cu
41.7 mol Cu
63.5 mol Cu

Answer

A

1. 1.04 mol Cu

Question 4

Q

01.05 The Mole Concept

How many moles are in a sample of 1.52 x 1024 atoms of mercury (Hg)? (3 points)

1.75 mol Hg
2.52 mol Hg
91.5 mol Hg
50.4 mol Hg

Answer

A

2. 2.52 mol Hg

Question 5

Q

01.05 The Mole Concept

Avogadro’s number is used to determine the number of subatomic particles in an atom. (2 points)

True

False

Question 6

Q

01.05 The Mole Concept

Which of the following processes will determine the number of moles in a sample? (3 points)

Dividing the mass of the sample by Avogadro’s number
Multiplying the mass of the sample by Avogadro’s number
Dividing the number of molecules in the sample by Avogadro’s number
Multiplying the number of molecules in the sample by Avogadro’s number

Answer

A

3. Dividing the number of molecules in the sample by Avogadro’s number

Question 7

Q

01.05 The Mole Concept

How many moles of silver are equivalent to 2.408 × 1024 atoms? (3 points)

4
2.5
0.25
0.4

Question 8

Q

01.05 The Mole Concept

What is the number of atoms in a mole of any element? (3 points)

Avogadro’s number
Graham’s number
Its atomic number
Its mass number

Answer

A

1. Avogadro’s number

Question 9

Q

01.05 The Mole Concept

What is the mass of 6.12 moles of arsenic (As)? (3 points)

12.2 g As
73.7 g As
276 g As
459 g As

Answer

A

4. 459 g As

Question 10

Q

01.05 The Mole Concept

How many atoms of iron (Fe) are in a sample of 7.38 mol Fe? (3 points)

1.23 x 10²³ atoms Fe
5.58 x 10²³ atoms Fe
3.37 x 10²⁴ atoms Fe
4.44 x 10²⁴ atoms Fe

Answer

A

4. 4.44 x 10²⁴ atoms Fe

Question 11

Q

01.05 The Mole Concept

Which of the following processes will determine the number of atoms in a sample? (3 points)

Dividing the mass of the sample by its molar mass
Multiplying the mass of the sample by its molar mass
Dividing the number of moles of the sample by Avogadro’s number
Multiplying the number of moles of the sample by Avogadro’s number

Answer

A

4. Multiplying the number of moles of the sample by Avogadro’s number

Question 12

Q

01.05 The Mole Concept

How many moles of gold are equivalent to 1.204 × 1024 atoms? (3 points)

0.2
0.5
2
5

Question 13

Q

01.05 The Mole Concept

What is the empirical formula of C6H12O6? (4 points)

C2H4O2

CH2O

CH2O2

C2H4O

Question 14

Q

01.05 The Mole Concept

What is a mole? What is the purpose?

Answer

A

Moles (mol): represent number & counting units

number of atoms in carbon 12 → avogadro’s number

Molar mass: total mass in grams on one mole in a substance

Amu on periodic table = 1 mole

Why use the mole?

many chemical properties depend on particle number (not mass)
Atoms too small count individually without special equipment
relationship between mass and number = determine numbe particles

Question 15

Q

01.05 The Mole Concept

What is Avogadro’s Number?

Answer

A

Number particles in mole experimentally determined variety of ways

Use: mass spectrometer (count atoms precisely)

Avogadro’s number: 6.02214 × 10²³

1 mole = 6.022 × 10²³particles

Question 16

Q

01.05 The Mole Concept

What is the relationship between moles and molar mass?

Answer

A

Mass of 1 Mole = Average Atomic Mass = Molar Mass

Question 17

Q

01.05 The Mole Concept

How to convert from number of moles to number of atoms:

Answer

A

Given number of atoms x (6.022 x 10²³ atoms) / 1 mole = number of atoms

Question 18

Q

01.05 The Mole Concept

How to convert from the number of atoms to moles?

Answer

A

Given number of atoms x (1 mole / 6.022 x 10²³ atoms) = number of moles

Question 19

Q

01.05 The Mole Concept

How to convert from grams to atoms:

Question 20

Q

How to convert from atoms to mass (g)?

Question 21

Q

05.02 Molar Mass of Compounds

Define mass value on the periodic table:

Answer

A

mass (grams) of one mole (6.022 * 10²³) of an element

Question 22

Q

05.02 Molar Mass of Compounds

How to calculate thee molar mass of a compound:

Answer

A

Subscrips: indicate amount of moles in compound

Step 01: Find molar mass on table

Step 02: multiply mass of each element by its subscript

Step 03: add up results

Question 23

Q

05.02 Molar Mass of Compounds

What is the molar mass of Na₂CO₃? (3 points)

60.0 g/mol
106.0 g/mol
118.0 g/mol
141.0 g/mol

Answer

A

2. 106.0 g/mol

Question 24

Q

05.02 Molar Mass of Compounds

How many moles of each element are in one mole of Be(OH)₂? (3 points)

1 mole of beryllium, 1 mole of oxygen, 2 moles of hydrogen
1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen
2 moles of beryllium, 2 moles of oxygen, 2 moles of hydrogen
2 moles of beryllium, 1 mole of oxygen, 1 moles of hydrogen

Answer

A

2. 1 mole of beryllium, 2 moles of oxygen, 2 moles of hydrogen

Question 25

Q

05.02 Molar Mass of Compounds

How many moles are in 123.0 grams of KClO₄? (3 points)

0.2354 mol KClO₄
0.6445 mol KClO₄
0.7124 mol KClO₄
0.8878 mol KClO₄

Answer

A

4. 0.8878 mol KClO₄

Question 26

Q

05.02 Molar Mass of Compounds

What is the mass of 4.32 mol GaI₃? (3 points)

455 g
1,230 g
1,720 g
1,950 g

Answer

A

4. 1,950 g

Question 27

Q

05.02 Molar Mass of Compounds

What is the molar mass of H₂SO₄?

(Molar mass of H = 1.0079 g/mol; S = 32.065 g/mol; O = 15.999 g/mol) (3 points)

98.08 g/mol
88.22 g/mol
86.79 g/mol
79.02 g/mol

Answer

A

1. 98.08 g/mol

Question 28

Q

05.02 Molar Mass of Compounds

Calculate the mass of 3.4 moles of nitric acid (HNO₃). Explain the process or show your work by including all values used to determine the answer. (5 points)

Answer

A

Firstly, determine the molar mass of nitric acid. There is 1 atom of Hydrogen, 1 atom of Nitrogen, and three atoms of Oxygen. Determine their individual masses and add the multiple of each other to find the total molar mass of the compound:

1H+1N+3O=1(1.0079)+1(14.0067)+3(15.9994)=63.0128

Therefore, the molar mass is 63.0128 g/mol HNO3.

Next, in order to determine the mass of 3.4 of HNO3, an appropriate conversion factor needs to be found. Since the conversion is from moles to mass, the following formula is needed: Given moles of a sample * (molar mass of compound) ÷ (1 mole) = mass of a sample. Hence, the conversion is 3.4 mol * (63.0128 g/mol)/1 mol = x.

3.4⋅63.01281=214.24g. Therefore, the mass of 3.4 moles of nitric acid is 214.24 grams.

Question 29

Q

05.03 The Empirical Formula

What is Percent Composition?

Answer

A

Ratios
represented empirical formula & molar mass

Step 01: Find the number of atoms in the empirical formula

Step 02: Find the molar mass

find on periodic table
multiply value in empirical formula

Step 03: Total individual molar masses to find total

Step 04: Calculate Percentage Composition by divide the molar mass of each element by molar mass of total molar mass

Question 30

Q

05.03 The Empirical Formula

Determine Empirical Formula:

Answer

A

Step 01: Check composition of compound (should be in grams)

if the sample is in percentage convert it assuming it is a 100-gram sample

Step 02: Convert amount element from grams to moles

conversion factor of (1 mole) ÷ (molar mass of substance)

Step 03: Divide each mole value by lowest mole value of all elements = whole-number mole ratio

mole values are mole ratio
empirical formula requires lowest whole-number ratio

Step 04: If the values are not a whole number, multiply all mole values by a number make them a whole number

Step 05: Use whole-numbers as subscripts when writing the empirical formula

Question 31

Q

05.03 The Empirical Formula

How to convert from the Empirical formula to the Molecular formula?

Answer

A

Step 01: Calculate molar mass of empirical formula in same way you would calculate molar mass of a compound

Step 02: Divide (entire compound) molecular mass by mass empirical formula → ratio between molecular formula and empirical formula (whole number)

Step 03: Multiply subscripts in empirical formula by ratio to get molecular formula

Question 32

Q

05.03 The Empirical Formula

What is the empirical formula of C₆H₁₂O₆? (4 points)

C₂H₄O₂
CH₂O
CH₂O₂
C₂H₄O

Answer

A

2. CH₂O

Question 33

Q

05.03 The Empirical Formula

Which pair shares the same empirical formula? (4 points)

C2H4 and C6H6
C6H6 and C3H3
CH2 and C6H6
CH and C2H4

Answer

A

2. C₆H₆ and C₃H₃

Question 34

Q

05.03 The Empirical Formula

What is the percent composition of NaHCO₃? (4 points)

23.20% Na, 4.26% H, 18.41% C, and 54.13% O
24.21% Na, 4.35% H, 12.26% C, and 59.18% O
27.36% Na, 1.20% H, 14.30% C, and 57.14% O
30.44% Na, 2.12% H, 10.25% C, and 60.19% O

Answer

A

3. 27.36% Na, 1.20% H, 14.30% C, and 57.14% O

Question 35

Q

05.03 The Empirical Formula

A compound is 40% carbon, 6.7% hydrogen, and 53.3% oxygen. What is the empirical formula? (4 points)

CH₂O₂
CH₂O
C₂H₄O
C₂H₄O₂

Answer

A

2. CH₂O

Question 36

Q

05.03 The Empirical Formula

The empirical formula of a chemical substance is CH. The molar mass of a molecule of the substance is 78.11 g/mol. What is the molecular formula of the chemical substance? (4 points)

C2H2
C2H4
C3H4
C6H6

Question 37

Q

05.04 Stoichiometry

Define stoichiometry:

Answer

A

dimensional analysis (math conversions) starts with one substance and ends with different substance

Question 38

Q

05.04 Stoichiometry

What is a mole ratio?

Answer

A

a conversion factor used in stoichiometry that describes the ratio between compounds or elements in a chemical reaction

Question 39

Q

05.04 Stoichiometry

How do you determine the mass value of a substance?

Answer

A

Step 01: Write a balanced Equation

Step 02: Begin stoichiometry using molar mass of starting substance

Step 03: Create mole ratio from coefficients of two compounds of interest

Step 04: Include mole ratio in stoichiometry conversion

Step 05: Include conversion factor for the molar mass of ending substance

Step 06: Solve through dimensional analysis

Question 40

Q

05.04 Stoichiometry

What is a valid mole ratio from the balanced equation 2C₃H₆ + 9O₂ → 6CO₂ + 6H₂O? (4 points)

Answer

A

6 mole H20 / 9 mole O2

Question 41

Q

05.04 Stoichiometry

How many moles of calcium chloride (CaCl2) are needed to react completely with 6.2 moles of silver nitrate (AgNO3)?

2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃₎₂ (4 points)

2.2 mol CaCl2
3.1 mol CaCl2
6.2 mol CaCl2
12.4 mol CaCl2

Answer

A

2. 3.1 mol CaCl2

Question 42

Q

05.04 Stoichiometry

What mass of Fe can be produced by the reaction of 75.0 g CO with excess Fe₂O₃ according to the equation: Fe2O₃(s) + CO(g) → Fe(s) + CO₂(g)? (4 points)

49.8 g Fe
99.7 g Fe
224 g Fe
299 g Fe

Answer

A

2. 99.7 g Fe

Question 43

Q

05.04 Stoichiometry

How many moles of sulfuric acid (H2SO4) are needed to react completely with 6.8 moles of lithium hydroxide (LiOH)?

2LiOH + H₂SO₄ → Li₂SO₄+ 2H₂O (4 points)

3.4 mol H2SO4
6.8 mol H2SO4
10.2 mol H2SO4
13.6 mol H2SO4

Answer

A

1. 3.4 mol H2SO4

Question 44

Q

05.04 Stoichiometry

What mass of Fe can be produced by the reaction of 75.0 g CO with excess Fe2O3 according to the equation: Fe2O3(s) + CO(g) → Fe(s) + CO2(g)? (4 points)

49.8 g Fe
99.7 g Fe
224 g Fe
299 g Fe

Answer

A

2. 99.7 g Fe

Question 45

Q

05.04 Stoichiometry

In an experiment, potassium chlorate decomposed according to the following chemical equation.

KClO₃ → KCl + O₂

(Molar mass of KClO3 = 122.5 g/mol; KCl = 74.55 g/mol; O2 = 31.998 g/mol)

If the mass of potassium chlorate was 240 grams, which of the following calculations can be used to determine the mass of oxygen gas formed? (4 points)

(240 × 2 × 31.998) ÷ (122.5 × 3) grams
(240 × 3 × 31.998) ÷ (122.5 × 2) grams
(240 × 2 × 122.5) ÷ (31.998 × 3) grams
(240 × 3 × 122.5) ÷ (31.998 × 2) grams

Answer

A

2. (240 × 3 × 31.998) ÷ (122.5 × 2) grams

Question 46

Q

05.05 Limiting Reactant

Define limiting and excess reactant:

Answer

A

Limiting Reactant: Runs out first

Excess Reactant: not used up completely

Question 47

Q

05.05 Limiting Reactant

Define Theoretical Yield:

Answer

A

maximum amount product expect from reaction (one run out before the other)

Question 48

Q

05.05 Limiting Reactant

What are the steps to determine the limiting reactants and moles?

Answer

A

Step 01: Determine limiting reactant by identifying the given about of reactants and what product is being solved for

Step 02: Write balanced equations and use it to create a mole ratio for the given reactant and product

Step 03: Set up and solve a stoichiometry calculation for each reactant using the mole ratios

Step 04: The Lower amous is the theoretical yield

Question 49

Q

05.05 Limiting Reactant

How to determine the mass of a theoretical yield?

Answer

A

Step 01: Determine if this is a limiting reactant problem by recognizing there are given starting amounts for both reactants. Balance the chemical reaction.

Step 02: Set up the stoichiometry calculation for each reactant using the periodic table to determine individual moral mass

Step 03: Use balanced equation to create mole ratio for the given reactants and products

Step 04: Include mole ratio for each stoichiometry calculation

Step 05: Include conversion factor for molar mass of ending substance for each calculation (solve through dimensional analysis)

Step 06: Lower amount is theoretical yield

Question 50

Q

05.05 Limiting Reactant

Consider the reaction 2CuCl₂ + 4KI → 2CuI + 4KCl + I2. If 4 moles of CuCl2 react with 4 moles of KI, what is the limiting reactant? (3 points)

CuCl2
KI
CuI
I2

Question 51

Q

05.05 Limiting Reactant

If 18.5 grams of CuCl2 react with 22.8 grams of NaNO3, what mass of NaCl can be formed?

CuCl₂ + NaNO₃ → Cu(NO₃)₂ + NaCl (3 points)

12.4 g NaCl
15.7 g NaCl
16.2 g NaCl
28.4 g NaCl

Answer

A

2. 15.7 g NaCl

Question 52

Q

05.05 Limiting Reactant

Read the chemical equation.

Fe₂O₃ + CO → Fe + CO₂

If 3 moles of Fe₂O₃react with 1.5 moles of CO, how many moles of each product are formed? (3 points)

1 mole of Fe and 1.5 moles of CO2
0.5 mole of Fe and 1 mole of CO2
6 moles of Fe and 9 moles of CO2
3 moles of Fe and 2 moles of CO2

Answer

A

1, 1 mole of Fe and 1.5 moles of CO2

Question 53

Q

05.06 Percent Yield

Define percentage yield:

Answer

A

Percent Yield: ^{Actual Yield} / _{Theoretical Yield} * 100

Percent Yield: ration acual yield to theoretical yield multiplied by 100

Theoretical Yield: maximum amount of product able to be produced from a given amount of reactant(s) (calculated)

Actual Yield: amount of product that is actually formed (measured)

reasons:

conditions not allowed reaction run to completion
purification of product results in loss of some of the product

The actual yield cannot exceed the theoretical yield. In practice, sometimes it appears that way, but that’s because an error occurred in the lab. This will be important to remember as you complete percent yield calculations.

Question 54

Q

05.09 Honors Stoichiometry Exam

Question 55

Q

05.09 Honors Stoichiometry Exam

How many atoms of Mg are present in 97.22 grams of Mg? (4 points)

6.022 × 10²³
2.408 × 10²⁴
4.818 × 10²⁴
5.855 × 10²⁵

Answer

A

2. 2.408 × 10²⁴

Question 56

Q

05.09 Honors Stoichiometry Exam

One mole of zinc has a mass of 65.4 grams. Approximately how many atoms of zinc are present in one mole of zinc? (4 points)

32 × 10²³ atoms
6 × 10²³ atoms
66 atoms
65 atoms

Answer

A

2. 6 × 10²³ atoms

Question 57

Q

05.09 Honors Stoichiometry Exam

Which of the following best defines the molar mass of a substance? (4 points)

Mass, in grams, of each particle of the substance
Total mass, in grams, of 100 particles of the substance
Mass, in grams, of the nucleus of the substance’s atoms
Total mass, in grams, of all the particles in one mole of the substance

Answer

A

4. Total mass, in grams, of all the particles in one mole of the substance

Question 58

Q

05.09 Honors Stoichiometry Exam

Which formula can be used to calculate the molar mass of ammonia (NH3)? (4 points)

molar mass of N + molar mass of H
3 × molar mass of N + molar mass of H
molar mass of N + 3 × molar mass of H
3 × molar mass of N + 3 × molar mass of H

Answer

A

3. molar mass of N + 3 × molar mass of H

Question 59

Q

05.09 Honors Stoichiometry Exam

How many moles of MgCO3 are present in 252.939 grams of MgCO3? (4 points)

2
3
5
6

Question 60

Q

05.09 Honors Stoichiometry Exam

Which pair of compounds has the same empirical formula? (4 points)

C2H6 and CH
CH3 and C2H6
C3H8 and C3H
C2H2 and C2H

Answer

A

2. CH3 and C2H6

Question 61

Q

05.09 Honors Stoichiometry Exam

What is the percentage composition of each element in hydrogen peroxide, H2O2? (4 points)

7.01% H and 92.99% O
7.22% H and 92.78% O
6.32% H and 93.68% O
5.88% H and 94.12% O

Answer

A

4. 5.88% H and 94.12% O

Question 62

Q

05.09 Honors Stoichiometry Exam

Determine the empirical formula of a compound containing 47.37 grams of carbon, 10.59 grams of hydrogen, and 42.04 grams of oxygen.

In an experiment, the molar mass of the compound was determined to be 228.276 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (8 points)

Answer

A

Empirical Formula:

Convert to moles:

37 grams of carbon = 3.944 moles of carbon
58 grams of hydrogen = 10.497 moles of hydrogen
04 grams of oxygen = 2.628 moles of oxygen

Divide each element by the lowest value:

944 mol C ÷ 2.628 = 1.500 mol C
497 mol H ÷ 2.628 = 3.994 = 4 mol H
628 mol O ÷ 2.628 = 1 mole O
5 * 2 = 3 mol C

4 * 2 = 8 mol H

1 * 2 = 2 mol O

Therefore, the empirical formula is C3H8O2.

Molecular Formula:

Molar mass of molecule: 228.276 g/mol

Molar mass empirical formula: 3(12.0107) + 8(1.0079) + 2(15.9994) = 76.0941 g/mol

=228.27676.0941

=2.999≈3

=3⋅C3H8O2

=C9H24O6

Question 63

Q

05.09 Honors Stoichiometry Exam

Which of the following is necessary to do a complete stoichiometric calculation? (4 points)

Write the mole ratio.
Add the nuclear masses of the products.
Divide the nuclear masses of the reactants.
Write the number of atoms in one mole of the product.

Answer

A

1. Write the mole ratio

Question 64

Q

05.09 Honors Stoichiometry Exam

The following reaction shows sodium hydroxide reacting with sulfuric acid.

4NaOH + 2H₂SO₄ → 2Na₂SO₄ + 4H₂O

How many grams of Na2SO4 are produced from 10.0 grams of NaOH?

(Molar mass of Na = 22.989 g/mol, O = 15.999 g/mol, H = 1.008 g/mol, S = 32.065 g/mol) (4 points)

17.8 grams
19.2 grams
35.5 grams
38.5 grams

Answer

A

1.

17.8 grams

Answer 55

A

3. 7.2 moles

Answer 56

A

3. The reactant that is used up before the others

Answer 57

A

1. 3.0 moles of water

Answer 58

A

2. About 0.20 grams of copper is formed, and some aluminum is left in the reaction mixture

Answer 59

A

4. The maximum amount of product that can be obtained

Answer 60

A

Theoretical Yield:

Since the chlorine gas in excess, sodium is the limiting reactant in the reaction. The balanced equation for the above reaction is 2Na+Cl2→2NaCl. The theoretical yield will be as follows:

=57.51⋅145.9794⋅22⋅116.88541

=146.172 grams theoretical yield

Actual Yield:

Actual Yield = (Percent Yield * Theoretical Yield) ÷ 100

=12570.792100

=125.708100

=125.708 grams actual yield