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12.1 Test Review
What is percent composition?
• the percent (by mass) of all the elements in a compound
What is Avagadros’ constant – 6.02 X 1023
• the number of particles contained in 1 mole of a substance
What is the SI unit of chemical quantity?
• the mole
What is molar mass?
• the mass of 1 mole of a substance
classify the following substance as:
a. formula unit b. atomc. molecule
d. ion•Mg
•atom
classify the following substance as:
a. formula unit b. atomc. molecule
d. ion•C2O4
-2
•ion
classify the following substance as:
a. formula unit b. atomc. molecule
d. ion• Na(NO3)2
•formula unit
classify the following substance as:
a. formula unit b. atomc. molecule
d. ion• C6H12O6
•molecule
determine the molar mass for the following substance:
• Br
•80
determine the molar mass for the following substance:
• C6H12O6
•180
determine the molar mass for the following substance:
• Na(NO3)2
•147
Simplified mole conversion
How many moles are in 32 grams of SCl2?
32g SCl2 X
1
1 mole SCl2102 g SCl2
= 0.31 mol SCl2
Calculate the number of atoms in 3.2 moles of gold (Au)
3.2g mol Au X
1
6.02 X 10 23 atoms Au 1 mol Au
= 1.9 X 1024 atoms Au
What is the volume in liters of 12.4 mol of CO2 gas?
12.4 mol CO2 X
1
22.4 L CO2
1 mol CO2
= 277.8 L CO2
What is the volume in liters of 21.3g of N2 gas?
21.3 g N X
1
X 22.4 L N2
1 mol N2
= 17.04 L N2
1 mol N2
28 g N2
How many molecules are in 72.3 L of O2?
72.3 L O2 X
1
X 6.02 X 1023 O2
1 mol O2
= 1.9 X1024 molecules O2
1 mol O2
22.4 L O2
What is the mass in grams of 2.1 X 1021 formula units of MgCl2?
X 94 g MgCl2
1 mol MgCl2
2.1 X 1021 fu MgCl2 1
= 0.33 g MgCl2
X 1 mol MgCl2
6.02 X 1023 fu MgCl2
Determine the percent composition for the following compound: CaCl2
• Ca = 40.078 g• Cl2 = 2 x 35.453 g = + 70.906 g • CaCl2 = 110.984 g
Ca = 40.1g X 100 = 36.1% Ca 110.9Cl = 70.9g X 100 = 63.9% Cl 110.9g
empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O
38.4 g Mn X 1mol Mn = 0.699 mole Mn 1 55 g Mn
16.8 g C X 1mol C = 1.339 mole C 1 12 g C
45.7 g O X 1mol O = 2.798 mole O 1 16 g O
0.699 mole Mn = 1 mol Mn0.699
1.339 mole C = 2 mol C 0.699
2.798 mole O = 4 mol O 0.699
So formula is MnC2O4
empirical and molecular formulas of a compoundwith percent composition of 71.0 % Cl, 25.0% C, 4.0 % H and mass 98.96 g/mol
71.0 g Cl X 1mol Cl = 2.0285 mole Cl 1 35 g Cl
25.0 g C X 1mol C = 2.0833 mole C 1 12 g C
4.0 g H X 1mol H = 4.0 mole H 1 1 g H
2.0285 mole Cl = 1 mol Cl 2.0285
2.0833 mole C = 1.02 mol C 2.0285
4.0 mole H = 1.97 mol H 2.0285
So formula is ClCH2
molecular mass of compound 98.96g = 2.01
molar mass of ClCH2 49 g
2 X (ClCH2) = Cl2C2H4