Chemistry 2202 Unit I Section 1 Homework Portfolio- KEY 2004-2005Page 1 of 8

Chemistry 2202 Unit I Section 1 Homework Portfolio- KEY 2004-2005Page 1 of 8

CHEMISTRY 2202

Unit I Section 1 Homework Portfolio - KEY

2004B 2005

(Total Value: 2 marks)

Value

(4)1.Copy and complete this table.

Isotope
Notation / Isotope
Name / Atomic Number / Mass
Number / Number of Protons / Number of
Neutrons / Number of
Electrons
/ Strontium-87 / 38 / 87 / 38 / 49 / 38
/ Thorium-230 / 90 / 230 / 90 / 140 / 90
/ Technetium-99 / 43 / 99 / 43 / 56 / 43
/ Lead-208 / 82 / 208 / 82 / 126 / 82

(2)2. One common type of smoke detector contains a small amount of radioactive americium-241. When smoke enters the detector the radiation inside the detector ionizes the smoke. The ionized smoke then conducts electricity and the alarm buzzes.

a) Write americium-241 in isotope notation?

b) How many neutrons does this isotope have? 146 neutrons

(4)3.Calculate the molar mass of each substance.

Chemistry 2202 Unit I Section 1 Homework Portfolio- KEY 2004-2005Page 1 of 8

(a)methanol

(b)nitrogen gas

(c)nickel (III) bromide

(e)calcium chloride hexahydrate

Chemistry 2202 Unit I Section 1 Homework Portfolio- KEY 2004-2005Page 1 of 8

(6)3.Calculate the mass of each given amount.

(a)2.50 mol of methane, CH4

Given: n = 2.50 mol of methane

Find: m = ? g

Required: Molar mass CH4

1C 12.01 g/mol = 12.01 g/mol

4 H  1.01 g/mol = 4.04 g/mol

16.05 g/mol

m = n X M = 2.50 mol x 16.05 g/mol = 40.125 g = 40.1 g CH4

(b)0.500 mol of sodium silcate, Na2SiO3

Given: n = 0.500 mol of sodium silcate

Find: m = ? g

Required: Molar mass Na2SiO3

2 Na X 22.99 g/mol = 45.98 g/mol

1 Si X 28.09 g/mol = 28.09 g/mol

3 O X 16.00 g/mol = 48.00 g/mol

122.07 g/mol

m = n X M = 0.500 mol X 122.07 g/mol = 61.035 g = 61.0 g Na2SiO3

Given: n = 0.250 mol of copper(II) nitrate dihydrate

Find: m = ? g

Required: Molar mass

1 Cu x 63.55 g/mol = 63.55 g/mol

2 N x 14.01 g/mol = 28.02 g/mol

6 Ox 16.00 g/mol = 96.00 g/mol

2 H2O x 18.02 g/mol = 36.04 g/mol

223.61 g/mol

m = n x M = 0.250 mol x 223.61 g/mol = 55.9025 g = 55.9 g

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

(6)4.Calculate the number of moles in each sample.

(a)100.0 g of gold metal, Au

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Given: 100.0 g Au

Find : n = ? mol

Required information: M = 196.97 g/mol

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

(b)50.0 g of silicon dioxide (sand), SiO2

Given: m = 50.0 g of SiO2

Find: n = ? mol

Required:

M = ? g/mol = 1(28.09) + 2(16.00) = 60.09 g/mol

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Given: m = 1.50 kg of KNO3 (convert to g: 1500 g)

Find: n = ? mol

Required:

M = ? g/mol = 39.10 + 14.01 + 48.00 = 101.11 g/mol

(2)5.Calculate the volume of each gas at STP.

(a)5.00 mol of oxygen gas

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Given: n = 5.00 mol of oxygen O2

Find v = ? L

Required: V = 22.4 L/mol

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

b) 1.25kmol of carbon monoxide

Given: n = 1.25 kmol of carbon monoxide CO (convert 1250 mol)

Find v = ? L

Required: V = 22.4 L/mol

(2)6.Calculate the number of moles of each gas at STP conditions.

(a)50.0 L of butane, C4H10

Find: n = ? mol

Required: V = 22.4 L/mol

(b)200.0kL of ozone, O3

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Find: n = ? mol

Required: V = 22.4 L/mol

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

(2)7.Calculate the number of molecules in each sample.

(a)0.125 mol of hydrogen

(b)0.250 kmol of carbon dioxide

(2)8.Calculate the number of moles in each sample.

(a) 7.50 x 1024 formula units of potassium chloride

(b)1.25 x 1022 atoms of barium

(4)9.Perform these conversions:

(a)25.0 g of strontium fluoride to number of formula units.SrF2

Find: number of formula units

Required: M = 1(87.62) + 2(19.00) = 125.62 g/mol

(b)250 L of methane, CH4 at STP to a mass amount.

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Given: v = 250 L CH4

Find: m = ? g

Required: VSTP = 22.4 L/mol

M = 1(12.01) + 4(1.01) = 16.04 g/mol

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

(3)10.Calculate the percentage composition of the elements in benzene, C6H6.

Required: M = 6(12.01) + 6(1.01) = 78.12 g/mol

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Calculate percentage C:

Calculate percentage H:

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

(4)11.The fermentation of apples produces apple cider. Upon chemical analysis of the cider a chemical compound is isolated and is found to be composed of 52.14% carbon (C), 13.13% hydrogen (H) and 34.73% oxygen (O). Based on this information, what is the empirical formula of this compound?

Given: 52.14% C, 13.13% H and 34.73% O

Find: Empirical Formula

Percent to mass (assume 100.0 g sample):

52.14g C 13.13g H34.73g O

Mass to Mole Conversion:

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

CarbonHydrogen Oxygen:

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Divide by Small:

CarbonHydrogenOxygen

Multiply until Whole: (already whole numbers!)

Write the empirical formula: C2H6O

(4)12.A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass and had a molar mass of 283.88 g/mol. What are the compound’s empirical and molecular formulas?

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Given: 43.64% P and 56.36% O

Find: Empirical Formula / molecular formula

Percent to mass (assume 100.0 g sample):

43.64g P56.36g O

Mass to Mole Conversion:

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

PhosporusOxygen:

Chemistry 2202 Unit I Section 1 Assignment 2003-2004-keyPage 1 of 8

Divide by Small:

PhosphorusOxygen

Multiply until Whole:

PhosphorusOxygen

2(1.00) = 2.002(2.50) = 5.00

Write the empirical formula: P2O5

Find ratio of empirical – molecular molar masses:

Mempirical = 2(30.97) + 5(16.00) = 141.94 g/mol

molecular formula = 2(empirical formula) = P4O10