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CH 10 THE MOLE QUIZ Multiple Choice Identify the choice that best completes the statement or answers the question. ____

1. Which is the correct formula for the compound whose percent composition is shown?

a. K2NO2 b. KNO2

c. KNO3 d. K4N10O3

____

2. Which element has a molar mass of 30.974 g/mol? a. Potassium c. Gallium b. Phosphorus d. Palladium

____

3. Which is the molar mass of the element Calcium? a. 20 g/mol c. 40.078 g/mol b. 40 g/mol d. 20.180 g/mol

____

4. Which is the correct empirical formula for this substance? Element Percent Composition C 37.5% H 12.5% O 50.00% a. CH4O b. CHO

c. C3H12O3 d. C3H4O

____

5. Which is the correct molar mass for the compound FeSO 4? a. 103.85 g/mol c. 415.4 g/mol b. 151.91 g/mol d. 247.85 g/mol

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6. Which is the correct molar mass for the compound CaBr 2? a. 120 g/mol c. 240 grams b. 200 mol d. 200 g/mol

____

7. Which is the mass of 8 moles of sodium chloride? a. 7.3 grams c. 0.137 g/mol b. 468 grams d. 468 mol

____

8. Which is the percent composition of bromine in the compound NaBr?

a. 81.6% b. 79.9% ____

c. 84.1% d. 77.7%

9. Which term is described as the percent by mass of any element in a compound? a. hydrate c. empirical formula b. molecular formula d. percent composition

____ 10. Which is the correct name for the hydrate whose composition is shown?

a. Iron (III) Phosphate Hydrate b. Iron (III) Hydrophosphate

c. Iron (III) Phosphate Tetrahydrate d. Iron (III) Phosphate Pentahydrate

____ 11. How many formula units are in 3.6 grams of NaCl? a. 0.06 c. 1.3  1026 21 b. 1.0  10 d. 3.7  1022 ____ 12. Which is the percent composition of phosphorus in Zn 3(PO4)2? a. 16.0 % c. 50.1 % b. 9.66 % d. 24.01 % ____ 13. How many grams are in 1.946 moles of NaCl? a. 113.7 g c. 0.033 g b. 30.1 g d. 44.7 g ____ 14. Which compound has a molar mass of 174.3 g/mol? a. K2SO4 c. MgCO3 b. Al(NO3)3 d. Ca(NO3)2 ____ 15. How many grams of phosphorus are in 500.0 grams of calcium phosphate? a. 49.92 grams c. 310.2 grams b. 99.89 grams d. 130.1 grams ____ 16. The empirical formula for a compound is CH2O, and the molar mass is 180.2 g/mol. Which is the molecular formula for this compound? a. C6H12O6 c. C8H20O4 b. C7H16O5 d. C3H6O3 ____ 17. Calculate the number of molecules in 2.5 moles of sucrose.

a. 0.15 x 1024 molecules b. 1.51 x 1024 molecules

c. 15.1 x 1024 molecules d. 150.5 x 1024 molecules

Short Answer 18. Find the formula of the compound with the percent composition shown.

19. Explain why the mole is used for counting in chemistry. 20. Compare and contrast an empirical formula and a molecular formula. 21. Find the number of moles in 9.035  1025 atoms of Boron. 22. Find the correct number of molecules of LiBr that are contained in a sample of 6.32 moles. 23. Find the empirical formula for the compound with this percent composition. Show your work. Element Percent Composition of Each Element Sodium 27.04 % Nitrogen 16.48% Oxygen 56.47% 24. List the steps needed to convert the number of atoms to the mass of an element. 25. How many molecules are there in 0.5 mol of sucrose?

26. A sample of calcium chloride, CaCl2, has a mass of 27.8 g.

a) How many moles of calcium chloride are in this mass? b) How many chloride ions are present? 27. Ascorbic acid, also known as Vitamin C, is composed of 40.9% carbon, 4.6% hydrogen, and 54.5% oxygen. The molar mass is 176 g/mol. Use the flow chart to help you calculate its molecular formula.

Problem 28. Determine the empirical formula for succinic acid that is composed of 40.60% carbon, 5.18% hydrogen, and 54.22% oxygen.

29. The atomic mass for boron is 10.811 g. How many molecules are in 320 g? 30. Arrange the quantities from largest to smallest in terms of mass: 1.35 moles of C, 2.51 x 10 24 molecules of Si, 150 g of K, and 2.5 moles of Zr. 31. An unknown compound contains 5.18 g of nitrogen, 17.74 g of oxygen, and 12.09 g of zinc. Use the flow chart to calculate the empirical formula for this compound.

CH 10 THE MOLE QUIZ Answer Section MULTIPLE CHOICE 1. ANS: NAT: 2. ANS: NAT: 3. ANS: NAT: 4. ANS: NAT: 5. ANS: NAT: 6. ANS: NAT: 7. ANS: NAT: 8. ANS: NAT: 9. ANS: NAT: 10. ANS: NAT: 11. ANS: NAT: 12. ANS: NAT:

C UCP.2 | B.2 B B.2 C B.2 A UCP.2 | B.2 B UCP.3 | B.2 D B.2 B B.2 D UCP.3 | B.2 D UCP.2 | B.2 C UCP.2 | B.2 D UCP.3 | B.2 A UCP.3 | B.3

PTS: 1

DIF: Bloom's Level 4|DOK2

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DIF: Bloom's Level 2|DOK2

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DIF: Bloom's Level 2|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 1|DOK1

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

13. ANS: NAT: 14. ANS: NAT: 15. ANS: NAT: 16. ANS: NAT: 17. ANS:

A UCP.3 | B.3 A UCP.3 | B.2 B UCP.3 | B.2 A UCP.2 | B.2 B

PTS: 1

DIF: Bloom's Level 3|DOK2

PTS: 1

DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

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DIF: Bloom's Level 3|DOK2

Use the conversion factor, , to determine the number of molecules in 2.5 moles of sucrose. To represent the correct number using scientific notation, the coefficient is between 1 and 10. PTS: 1

DIF: Bloom's Level 3|DOK2

SHORT ANSWER 18. ANS: Ca3(PO4)2 PTS: 1 DIF: Bloom's Level 4|DOK2 NAT: UCP.2 19. ANS: The mole is used for counting in chemistry because individual atoms and molecules are so small, a very large number of them are contained in a single gram of a substance. A mole allows the number of particles to be quantified while providing the chemist with a convenient number to use in calculations. PTS: 1 DIF: Bloom's Level 2|DOK1 NAT: UCP.2 20. ANS: Both empirical and molecular formulas show the elements in a compound in the correct ratios. An empirical formula shows the lowest whole number ratio, while a molecular formula shows the actual number of atoms of each element in a single molecule of that compound. PTS: 1 21. ANS:

DIF: Bloom's Level 4|DOK2

NAT: UCP.2

DIF: Bloom's Level 3|DOK2

NAT: UCP.2 | UCP.3

150.1 moles PTS: 1 22. ANS:

molecules 3.80  1024 molecules PTS: 1 23. ANS:

DIF: Bloom's Level 3|DOK2

NAT: UCP.2 | UCP.3

NaNO3 PTS: 1 DIF: Bloom's Level 3|DOK2 NAT: UCP.2 24. ANS: To convert the number of atoms to the mass of an element: a. Multiply the number of atoms by the inverse of Avogadro’s number as a conversion factor. b. Multiply the calculated number of moles of the element by the molar mass of the element. PTS: 1 DIF: Bloom's Level 1|DOK1 NAT: UCP.3 | B.1 TOP: Calculate the number of atoms of an element when given the number of moles of the element. KEY: Atoms to mass conversion MSC: 2 25. ANS: There are 6.02 x 1023 molecules in every mole.

= 3.01 x 1023 molecules

0.5 mol x

PTS: 1 DIF: Bloom's Level 3|DOK2 26. ANS: Use the figure to help solve the problem. a) Determine the molar mass of CaCl2. Then calculate the number of moles. molar mass = (40.1 + 2(35.5)) = 111.1 g/mol CaCl2

27.8 g CaCl2

= 0.25 mol CaCl2

b)Calculate the number of molecules present in 0.25 mol CaCl2

0.25 mol CaCl2

= 1.51 x 1023 molecules CaCl2

To calculate the number of Cl 1– ions, use the ratio from the chemical formula as the conversion factor.

1.51 x 1023 molecules CaCl2 = 3.0 x 1023 molecules Cl1– PTS: 1 DIF: Bloom's Level 3|DOK2 27. ANS: The compound’s molecular formula is C6H8O6.

moles of C (in 100 g) = 40.9 g C

= 3.41 mol C

moles of H (in 100 g) = 4.6 g H

= 4.60 mol H

moles of O (in 100 g) = 54.5 g O

= 3.41 mol O

Determine the mole ratio by dividing by 3.41. (1.00 mol C):(1.35 mol H):(1.00 mol O) This is the simplest mole ratio. Multiply the mole ratios by 3. (3 mol C):(4 mol H):(3 mol O) The empirical formula is C3H4O3. Calculate the empirical formula mass: C (3 12 g/mol) + H (4 1 g/mol) + O (3 16 g/mol) = 88 g/mol Divide the molar mass by the empirical formula mass to determine n =2 Multiple the subscripts by 2 to obtain the molecular formula: C6H8O6 PTS: 1

DIF: Bloom's Level 3|DOK2

PROBLEM 28. ANS: C2H3O2 PTS: 1 DIF: Bloom's Level 3|DOK2 NAT: B.2 TOP: Determine the empirical and molecular formulas for a compound from mass percent and actual mass data. KEY: Empirical formula MSC: 3 29. ANS: = 1.78 x 1025 molecules

x

PTS: 1 DIF: Bloom's Level 3|DOK2 30. ANS: Substances arranged from largest to smallest are: Zr, K, Si, C

Zr

2.5 mol x

= 228 g

Si 2.51 x 1024 molecules x

C 1.35 mol x PTS: 1 31. ANS:

x

= 16.2 g DIF: Bloom's Level 3|DOK2

= 117 g

Calculate the number of moles for each element.

N 5.18 g

= 0.37 mol

O 17.74 g

= 1.10 mol

Zn 12.09 g

= 0.18 mol

Determine the ratio by dividing by the smallest value.

N

=2

O

=6

Zn

=1

The empirical formula for this compound is ZnN2O6. PTS: 1

DIF: Bloom's Level 3|DOK2...