Conversion Factor Analysis practice
Write down and answer these questions:
1) Molar Mass of C6H12O6 = ____________
2) How many C atoms in 1.74 mol of C6H12O6?
3) 2.34 mol H2SO4 are needed as a reactant.
a) How many grams of H2SO4 are needed?
b) How many kilograms (kg) would that be?
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
MH = 1.008 g/mol
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
MH = 1.008 g/mol
MO = 16.00 g/mol
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
MH = 1.008 g/mol
MO = 16.00 g/mol
6(12.01 g/mol)
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
MH = 1.008 g/mol
MO = 16.00 g/mol
6(12.01 g/mol) + 12(1.008 g/mol)
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
MH = 1.008 g/mol
MO = 16.00 g/mol
6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00g/mol) =
Conversion Factor Analysis practice
1) Molar Mass of C6H12O6 = ____________
M=g/mol
see Periodic Table for molar mass of each atom
MC = 12.01 g/mol
MH = 1.008 g/mol
MO = 16.00 g/mol
6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00g/mol) =
180.156 g/mol
Conversion Factor Analysis practice
2) How many C atoms in 1.74 mol of C6H12O6
1.74 mol x 6.02x1023 atoms x 6 C atoms = 2.625x1023
1 mol
24 atoms
3) 2.34 mol H2SO4 are needed as a reactant.
a) How many grams of H2SO4 are needed?
b) How many kilograms (kg) would that be?
First, setup the calculation to convert from what
you have to what you need:
2.34 mol x ________g H2SO4 =
mol
3) 2.34 mol H2SO4 are needed as a reactant.
a) How many grams of H2SO4 are needed?
b) How many kilograms (kg) would that be?
Second, find the quantities needed to plug into
your calculation:
Find molar mass of H2SO4:
2(1.008g/mol) + 1(32.06g/mol) + 4(16.00g/mol)
= 98.076g/mol
3) 2.34 mol H2SO4 are needed as a reactant.
a) How many grams of H2SO4 are needed?
b) How many kilograms (kg) would that be?
Third, plug the quantities into the equation to
convert to the needed units:
2.34 mol x 98.076g H2SO4 = 229.50 g H2SO4
mol
3) 2.34 mol H2SO4 are needed as a reactant.
a) How many grams of H2SO4 are needed?
b) How many kilograms (kg) would that be?
Fourth, convert from grams to kilograms:
229.50g x 1 kg . = 0.2295 kg this is the answer to (b)
1,000g
4) Find the mass of 1 mole of MgCl2 .
5) What is the mass of 3.7 moles of MgCl2?
6) How many Cl atoms in 3.7 mol MgCl2?
4) Find the mass of 1 mole of MgCl2 .
Look at periodic table.
Mg 24.31g
Cl2
4) Find the mass of 1 mole of MgCl2 .
Look at periodic table.
Mg 24.31g
Cl2 70.90g
4) Find the mass of 1 mole of MgCl2 .
Look at periodic table.
Mg 24.31g
Cl2 70.90g
95.21g
5) What is the mass of 3.7 moles of MgCl2?
95.21g x 3.7 mol = 352.3g
1 mol
6) How many Cl atoms in 3.7 mol MgCl2?
Step 1: Find total # of molecules
3.7mol x 6.02x1023 MgCl2 = 2.23x1024 MgCl2
1 mol
6) How many Cl atoms in 3.7 mol MgCl2?
Step 1: Find total # of molecules
3.7mol x 6.02x1023 MgCl2 = 2.23x1024 MgCl2
1 mol
Step 2: since there are 2 Cl atoms in each MgCl2
molecule, multiply by 2
2.23x1024 MgCl2 x 2 Cl = 4.46x1024 Cl atoms
1 MgCl2
Conversion Factor Analysis
practice
How many moles in your name?
Chalk = Calcium Carbonate, CaCO3
Initial mass:________g Final mass:_________g
Mass of CaCO3 in your name:__________g
Molar mass of CaCO3:________g/mol
How many moles CaCO3 in your name?
How many molecules of CaCO3 in your name?