Chapter 10 Worksheet Examples
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Transcript of Chapter 10 Worksheet Examples
Chapter 10 Worksheet Examples
MOLAR MASSWorksheet:
Example: Calculate the molar mass (gram molecular weight) of a mole of iodine, I2.
I = 126.9(2) = 253.8 molg
From the mass off of the periodic table.
From the formula of the compound
Unit for molar mass
Example: Calculate the molar mass of a mole of aluminum sulfate (Al2(SO4)3).
Al = 26.98 (2) = 53.96S = 32.07 (3) = 96.21O = 16.00 (12) = 192.00---------------------------------------- add together = 342.17 mol
g
From the masses off of the periodic table.
From the formula of the compound
Unit for molar mass
MOLE AS A UNIT OF MASSWorksheet:
Example: What is the mass of 5.00 moles of water?
90.1 gH2O
18.02 g/mol
5.00 mol
Molar Mass Calculation
H = 1.01 (2) = 2.02O = 16.00 (1) = 16.00------------------------------ = 18.02 g/mol
From the problem
Molar mass
X
Example: What is the mass of 0.50 moles of calcium carbonate?
50.045 gCaCO3
100.09 g/mol
0.50 mol
Molar Mass Calculation
Ca = 40.08 (1) = 40.08C = 12.01 (1) = 12.01O = 16.00 (3) = 48.00------------------------------ = 100.09 g/mol
From the problem
Molar mass
X
Example: How many moles of calcium chloride are in 333 grams of calcium chloride
333 gCaCl2
110.98 g/mol
3.00 mol
Molar Mass Calculation
Ca = 40.08 (1) = 40.08Cl = 35.45 (2) = 70.9------------------------------ = 110.98 g/mol
From the problem
Molar mass
÷
AVOGADRO’S NUMBERWorksheet:
Example: How many molecules of water are there in 3.00 moles of water?
1.806x1024 mclH2O
6.02x1023 mcl/mol
3.00 mol
From the problem
X
The number of atoms/molecules in one mole is always 6.02x1023
Example: How many moles of neon are there in 2.408x1024 atoms of neon?
2.408x1024 atmNe
6.02x1023 atm/mol
3.99 mol
From the problem
The number of atoms/molecules in one mole is always 6.02x1023
÷
MOLAR VOLUME OF A GASWorksheet:
Example: What is the volume, in liters, of a 2.00 mole sample of methane (CH4) at STP?
44.8 LCH4
22.4 L/mol
2.00 molFrom the problem
The number of liters in one mole is always 22.4
X
Example: How many moles of ethane (C2H6) are there in 5.60 liters of ethane?
5.6 LC2H6
22.4 L/mol
0.25 mol
From the problem
The number of liters in one mole is always 22.4÷
MIXED MOLE PROBLEMSWorksheet:
We’ve done problems of one step. Moles to a unit or unit to moles. Now we can do multi step problems using this picture to help us see where we need to go next.
This picture is a summary of all of the problems we have done up to this point along with helpful hints to units and numbers. The better you understand this the easier mole conversions will be.
Molar mass is g/mol and you find grams from the periodic table.
Example: What would be the volume in liters of 40.36 grams of neon at STP?
40.36 gNe
20.18 g/mol
2.00 mol
÷
44.8 LNe
22.4 L/mol
2.00 mol
X
We’re not to liters yet…
From the problem
From the periodic table
The number of liters in one mole is always 22.4
Example: How many molecules would there be in 56 liters of carbon dioxide at STP?
56 LCO2
22.4 L/mol
2.5 mol
÷
1.505x1024 mclCO2
6.02x1023 mcl/mol
2.5 mol
X
We’re not to molecules yet…
The number of liters in one mole is always 22.4
From the problem
The number of atoms/molecules in one mole is always 6.02x1023
PERCENT COMPOSITION & FORMULAS
Worksheet:
Example: Calculate the percent composition of sodium hydrogen carbonate (NaHCO3)?
Molar Mass CalculationNa = 22.99 (1) = 22.99H = 1.01 (1) = 1.01C = 12.01 (1) = 12.01O = 16.00(3) = 48.00---------------------------------------------- = 84.01 g/mol
Step 1 – find molar mass of the compound.Step 2 – divide mass of each element by molar mass.Step 3 – multiply answers by 100%.
C%30.14
%10001.8401.12
H%20.1
%10001.8401.1
Na%37.27
%10001.8499.22
O%14.57
%10001.8400.48
Example: Which pair of molecules has the same empirical formula?
a) C2H4O2
c) NaCrO4
b) C6H12O6
d) Na2Cr2O7
CH2O CH2O
An empirical formula is similar to a reduced fraction.
Molecular formula
Empirical formula
These two can not be “reduced”.
Example: Calculate the empirical formula for a compound with 67.6% Hg, 10.8% S, 21.6% O?Step 1 – change percents to grams (assume we have 100 gram sample then percents are the number of grams).Step 2 – change the grams to moles.Step 3 – divide all answers by the smallest number (of the answers).Step 4 – the results become the subscripts for the formula.
134.034.0
59.200
6.67
molmol
molg
Hgg
134.034.0
07.32
8.10
molmolmolg
Sg
434.035.1
00.16
6.21
molmolmolg
Og
÷ ÷
Step 1
Step 2
Step 3
Step 4
HgSO4
Example: Find the molecular formula of ethylene glycol, which is used as antifreeze.
The molecular mass is 62g/mol and the empirical formula is CH3O?
Step 1 – Find the empirical formula mass (EFM)Step 2 – Divide the molar mass (from problem) by the EFM to get the multiplier.Step 3 – Use the multiplier to determine the subscripts by multiplying each subscript by the multiplier.
Step 1 EFM Calculation
C = 12.01 (1) = 12.01H = 1.01 (3) = 3.03O = 16.00(1) = 16.00---------------------------------------- =31.04 g/mol
201.3262
Step 2
MULTIPLIER
Step 3
C2H6O2