Ch. 3 Stoichiometry: Calculations with Chemical Formulas

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Ch. 3 Stoichiometry: Calculations with Chemical Formulas

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Ch. 3 Stoichiometry: Calculations with Chemical Formulas

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Ch. 3 Stoichiometry: Calculations with Chemical Formulas

- Atoms are neither created nor destroyed during any chemical reaction. Atoms are simply rearranged.

- The quantitative nature of chemical formulas and chemical reactions

- The chemical formulas on the left of the arrow that represent the starting substances
2H2 + O2 2H2O

Reactants

- The substances that are produced in the reaction and appear to the right of the arrow
2H2 + O2 2H2O

Products

- Because atoms are neither created nor destroyed in any reaction a chemical equation must have the same number of atoms of each element on either side of the arrow

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

- Rapid reactions that produce a flame.
- Most combustion reactions involve O2 as a reactant
- Form CO2 and H2O as products

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

- 2 or more substances react to form 1 product.

2Mg(s) + O2(g) 2MgO(s)

Mg=1 x 2=2 Mg= 1 x 2=2

O= 2 O=1 x 2 = 2

- 1 substance undergoes a reaction to produce 2 or more substances

CaCO3 (s) CaO (s) + CO2(g)

Ca=1 Ca=1

C=1 C=1

O=3 O=1+2=3

- Counting
- Mass
- Volume

- If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?

- Count: 1 dozen apples = 12 apples
- Mass: 1 dozen apples = 2.0 kg apples
- Volume: 1 dozen apples = 0.20 bushels
apples

Conversion Factors:

- 1 dozen2.0 k.g1 dozen
12 apples 1 dozen 0.20 bushels

- 0.50 bushel x 1 dozen x 2.0 kg =
0.20 bushel 1 dozen

= 5.0 kg

- Named after the Italian scientist Amedo Avogadro di Quaregna
- 6.02 x 10 23

- 1 mol = 6.02 x 10 23 representative particles
- Representative particles: atoms, molecules ions, or formula units (ionic compound)

- Moles= representative x 1 mol
particles 6.02 x 10 23

- How many moles is 2.80 x 10 24 atoms of silicon?

- 2.80 x 10 24 atoms Si x 1 mol Si
6.02 x 10 23 atoms Si

= 4.65 mol Si

- How many molecules of water is 0.360 moles?

- 0.360 mol H2O x 6.02 x 10 23 molecules H2O1 mol H2O
=2.17 molecules H2O

- The atomic mass of an element expressed in grams = 1 mol of that element = molar mass

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

- If you have 4.5 mols of sodium, how much does it weigh?

- .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na
1 mol Na

- If you have 34.3 g of Iron, how many atoms are present?

- 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms
55.8 g Fe 1 mol Fe

=3.70 x 10 23 atoms Fe

- To find the mass of a mole of a compound you must know the formula of the compound
- H2O H= 1 g x 2
O= 16 g

18 g = 1 mole = 6.02 x 10 23

molecules

- What is the mass of 1 mole of sodium hydrogen carbonate?

- Sodium Hydrogen Carbonate = NaHCO3
- Na=23 g
- H=1 g
- C=12 g
- O=16 g x3
- 84 g NaHCO3 = 1 mol NaHCO3

- Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical

- States that equal volumes of gases at the same temperature and pressure contain the same number of particles
- Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible

- Volume of a gas changes depending on temperature and pressure
- STP= 0oC (273 K)
101.3 kPa (1 atm)

- At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume

- AT STP
- 1 mol gas22.4 L gas
22.4 L gas 1 mol gas

- At STP, what volume does 1.25 mol He occupy?

- 1.25 mol He x 22.4 L He = 28.0 L He
1 mol He

- If a tank contains 100. L of O2 gas, how many moles are present?

- 100. L O2 X 1 mol O2 = 4.46 mol O2
22.4 L O2

- The density of a gas at STP is measured in g/L
- This value can be sued to determine the molar mass of gas present

- A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.

- 1 mol gas x 22.4 L gas X 3.58 g gas =
1 mol gas 1 L gas

Molar Mass= 80.2 g

- The relative amounts of the elements in a compound
- These percentages must equal 100

- %element = mass of element x 100
mass of compound

- Find the mass percentage of each element present in Al2 (CO3)3

- Al2(CO3)3
- Al= 27 g x 2 = 54 g / 234 g x 100=23%
- C= 12 g x 3 = 36 g/ 234 g x 100= 15%
- O = 16 g x 9 = 144 g / 234 g x 100=62%
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

- Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical fromula.

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

- The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula
- We can obtain the molecular formula from the empirical formula IF we know the molecular weight of the compound.

- The empirical formula of ascorbic acid is C3H4O3. The molecular weight of ascorbic acid is 176 amu. Determine the molecular formula.

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

- The reactant that is completely consumed
- It determines or limits the amount of product that forms
- The other reactant(s) are called excess reagents

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?

- 2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
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

- The quantity of the product that is calculated to form when all of the limiting reactant reacts.

- 2C4H10(l) + 13 O2(g) 8CO2(g) + 10H2O(l)
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?

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)