Ch. 3 Stoichiometry: Calculations with Chemical Formulas. Law of Conservation of Mass. Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged. Stoichiometry. The quantitative nature of chemical formulas and chemical reactions. Reactants.
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Ch. 3 Stoichiometry: Calculations with Chemical Formulas
2H2 + O2 2H2O
Reactants
2H2 + O2 2H2O
Products
CH4 + O2 CO2 + H2O
C=1 C=1
H=4 H=2
O=2 O=3
CH4 + O2 CO2 + 2H2O
C=1 C=1
H=4 H=2 X 2 =4
O=2 O=3
CH4 + O2 CO2 + 2H2O
C=1 C=1
H=4 H=2 x 2 = 4
O=2 O= 2 + 2x1 = 4
CH4 + 2O2 CO2 + 2H2O
C=1 C=1
H=4 H=2 x 2 = 4
O=2 x 2 = 4 O= 2 + 2x1 = 4
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)
C= 3 C=1 X 3 = 3
H=8 H=2 X 4 = 8
O= 2 X 5 = 10 O=(2 X3)+(1X4)=10
2Mg(s) + O2(g) 2MgO(s)
Mg=1 x 2=2 Mg= 1 x 2=2
O= 2 O=1 x 2 = 2
CaCO3 (s) CaO (s) + CO2(g)
Ca=1 Ca=1
C=1 C=1
O=3 O=1+2=3
apples
Conversion Factors:
12 apples 1 dozen 0.20 bushels
0.20 bushel 1 dozen
= 5.0 kg
particles 6.02 x 10 23
6.02 x 10 23 atoms Si
= 4.65 mol Si
=2.17 molecules H2O
Molar mass S
Molar mass Hg
Molar mass C
Molar mass Fe
6.02 x 10 23 atoms S
6.02 x 10 23 atoms Hg
6.02 x 10 23 atoms C
6.02 x 10 23 atoms Fe
1 mol Na
55.8 g Fe 1 mol Fe
=3.70 x 10 23 atoms Fe
O= 16 g
18 g = 1 mole = 6.02 x 10 23
molecules
101.3 kPa (1 atm)
22.4 L gas 1 mol gas
1 mol He
22.4 L O2
1 mol gas 1 L gas
Molar Mass= 80.2 g
mass of compound
234 gAl2(CO3)3
The simplest whole number ratio of atoms in a compound
The formula obtained from percentage composition
Ex CH , CH4, H2O, C3H8
NOT C2H4, or C6H12O6 these could be simplified
Assume that you have 100 grams of the compound therefore:
Hg = 73.9 % 73.9 g
Cl= 26.1% 26.1 g
Step 2: Change grams of your compound to moles
Hg = 73.9 g x 1 mol =0.368 mol Hg
200.6g
Cl= 26.1 g x 1 mol = 0.735 mol Cl
35.5 g
Step 3: Find the lowest number of moles present
Hg = 73.9 g x 1 mol =0.368 mol Hg
200.6g
Cl= 26.1 g x 1 mol = 0.735 mol Cl
35.5 g
0.368 < 0.735
Step 4: Divide by the lowest number of moles to obtain whole numbers
Hg = 73.9 g x 1 mol = 0.368 mol = 1
200.6g 0.368 mol
Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2
35.5 g 0.368 mol
Step 5: Put the whole numbers into the empirical formula
Hg = 73.9 g x 1 mol = 0.368 mol = 1
200.6g 0.368 mol
Cl= 26.1 g x 1 mol = 0.735 mol= 1.99=2
35.5 g 0.368 mol
HgCl2
Step 1: First determine the molecular weight of the empirical formula
C3H4O3
C= 12 amu x 3
H= 1 amu x 4
O= 16 amu x 3
88 amu
Step 2: Divide the molecular weight of the molecular formula by the molecular weight of the empirical formula
C3H4O3176 amu = 2
C= 12 amu x 3 88 amu
H= 1 amu x 4
O= 16 amu x 3
88 amu
Step 3: Multiply the empirical formula by the number calculated in step 2
176 amu = 2
88 amu
(C3H4O3) x 2 = C6H8O6
2 H2 (g) + O2 (g) 2 H2O (l)
2 molecules 1 molecule2 molecules
Or since we can’t count out 2 molecules
2 mol 1 mol 2 moles
The coefficients in a chemical reaction can be interpreted as either the relative number of molecules (formula units) involved in the reaction OR the relative number of moles
2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
How many moles of O2 do you need to react with 5 moles of C4H10?
2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
How many moles of O2 do you need to react with 5 moles of C4H10?
5 mol C4H10 x 13 mol O2 = 32.5 mol O2
2 mol C4H10
2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
How many grams of O2 do you need to react with 50.0 g of C4H10?
2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
How many grams of O2 do you need to react with 50.0 g of C4H10?
50.0 g C4H10 x 1 mol C4H10 x 13 mol O2 x 32 g O2=179 g O2
58 g C4H10 2 mol C4H10 1 mol O2
2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
If you have 25.0 g O2 and 25.0 g of C4H10, what is the limiting reactant?
25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2
58 g C4H10 2 mol C4H10
25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2
32 g O2 13 mol O2
0.481 mol < 1.72 mol C4H10 is the limiting reactant
25.0 g C4H10 x 1 mol C4H10 x 8mol CO2 = 1.72 mol CO2
58 g C4H10 2 mol C4H10
25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2
32 g O2 13 mol O2
0.481 mol < 1.72 mol C4H10 is the limiting reactant Calculate the theoretical yeild
25.0 g C4H10 x 1 mol C4H10 x 8 mol CO2 = 0.481 mol CO2
32 g O2 13 mol O2
0.481 mol CO2 X 44 g CO2= 21.2 g CO2
1 mol CO2
If all of the limiting reactant (25.0 g C4H10) reacts than 21.2 g of CO2 will form.
Percent Yield = Actual Yield X 100
Theoretical yield
A student calculates that they should theoretically make 105 g of iron in an experiment. When they perform the experiment only 87.9 g of iron were produced. What is the percent yield?
87.9 g / 105 g x 100 = 83.7%
(actual) (theoretical)