Stoichiometry

• Let’s pretend we are going to a party and we want to bring marshmallow treats. Here is our recipe.
• 3 T of butter
• 40 marshmallows
• 6 c of rice cereal
• This will make 24 treats
• We realize that we only have 4.5 cups of cereal. What do we do?
• 4.5 c cereal x 40 marshmallows/6 c cereal
• Would this be enough to feed the 20 people that are coming?
• 4.5 c cereal x 24 treats/6 c cereal = 18 treats
• We use this same process to determine how much of the products we need in chemistry to make a certain amount of reactant.
• For example, in the equation P₄ + 5O₂ —> 2P₂O₅, we know that for every 1 molecule of of P₄ and 5 molecules of O_2, we are going to end up with 2 molecules of P₂O₅. 
• This gives us various relationships in our equation:
• 1 mol P₄ reacts to form 2 mol P₂O₅ (1:2 ratio)
• 1 mol P₄ reacts with 5 mol O₂ (1:5 ratio) and
• 5 mol O₂ reacts to form 2 mol P₂O₅ (5:2 ratio)
• Even if we were to cut the original quantities in half or double them, we would still have the same ratio. These are known as mole ratios.
• We can use these ratios to do mole-to-mole conversions. This would be to determine how much of a product we would get if we added a certain amount of reactants.
• If a chemical engineer wanted to produce 25.0 mol of diphosphorus pentoxide, how many moles of phosphors would they need?
• 25.0 mol P₂O₅ x 1 mol P₄/2 mol P₂O₅ = 12.5 mol P₄
• What is the amount of oxygen needed?
• 12.5 mol P₄ x 5 mol O₂/1 mol P₄ = 62.5 mol O₂
• If 25.0 mol of diphosphorus pentoxide reacts with water to form phosphoric acid, how many moles of water is required? P₂O₅ + 3H₂O —> 2H₃PO₄ 
• 25.00 mol P₂O₅ x 3 mol H₂O/ 1 mol P₂O₅ = 75.0 mol H₂O
• Why must the conversions be done in moles? Because the recipe/balanced equation is based on moles, not mass. 
• If we wanted to know the moles of diphosphorus pentoxide that would result from the burning of 1.55 kg of phosphorus from the previous equation, we first have to convert mass to moles.
• 1550 g P₄ x 1 mol/123.9 g P₄ = 12.5 mol P₄
• Then we can use our ratio to determine the diphosphorus pentoxide that would result:
• 12.5 mol P₄ x 2 mol P₂O₅/1 mol p ₄ = 25.0 mol P₂O₅
• Example: How many moles of phosphoric acid can be formed from 3550 g of diphosphorus pentoxide? P₂O₅ + 3H₂O —> 2H₃PO₄
• First convert 3550 g P₂O₅ to moles.
• 3550 g P₂O₅ x 1 mol P₂O₅/141.9 g P₂O₅ = 25.0 mol P₂O₅ 
• Then use ratio to figure out the rest
• 25.0 mol P₂O₅ x 2 mol H₃PO₄/1 mol P₂O₅ = 50.0 mol H₃PO₄
• What if we want need to figure out mass given mass? 
• What mass of water will react with 3500 g of diphosphorus pentoxide? P₂O₅ + 3H₂O —> 2H₃PO₄
• 3550 g P₂O₅ x 1 mol P₂O₅/141.9 g P₂O₅ = 25.0 mol P₂O₅ 
• 25.0 mol P₂O₅ x 3 mol H₂O/1 mol P₂O₅ = 75.0 mol H₂O
• 75.0 mol H₂O x 18.02 g H₂O/1 mol H₂O = 1350 g H₂O

Calculating Empirical Formula from % Composition

• Empirical formulas contain the information needed to calculate the percent composition of compounds since they preserve the ration of atoms in a compound. When the mole ratio is conserved, so is the mass ratio. 
• Scientists use this to determine the empirical formula of a compound by experimentally determining the percent composition. 
• Suppose a chemist is given 100.0 g of an unknown substance. After analysis, he concludes that it contains 75.00 g (75%) C and 25.00 g (25%) O. We are able to figure out the empirical formula from this information.
• 75.00 g C x 1 mol C/12.01 g C = 6.245 mol C
• 25.00 g H x 1 mol H/1.01 g H = 24.80 mol H
• Next to get a 1: something ratio, we divide both numbers by the smallest.
• 6.245/6.245 = 1
• 24.80/6.245 = 3.97 (this gets rounded to 4)
• with a ratio of 1:4, we can write our formula as C₁H₄ or CH₄
• A laboratory analysis of an unknown gas has determined that the bass is 72.55% oxygen and 27.45% carbon by mass. What is the empirical formula of the compound?
• 73.55 g O x 1 mol O/16.00 g O = 4.534 mol O
• 27.45 g C x 1 mol g C/12.01 g C = 2.286 mol C
• 4.543/2.286 : 2.286/2.286 = 1.983:1.000
• Giving a ratio of 2 O:1 C
• CO₂
• A 5.000 g sample of an unknown compound contains 1.844 g of N and 3.156 g of O. Find the empirical formula.
• 1.844 g N x 1 mol N/14.01 g N = 0.1316 mol N
• 3.156 g O x 1 mol O/16.00 g O = 0.1973 mol O
• 0.1316/0.1316 : 0.1973/0.1316 = 1:1.499
• Because we do not have half numbers in empirical formulas, we must eliminate the fraction. Making this 2:3
• N₂O₃
• Caffeine, in coffee and some carbonated beverages, is 5.170% H, 16.49% O, 28.86% N, and 49.48% C by mass. What is the empirical formula of caffeine? (C₄H₅N₂O)

Practice for Test

A. Find moles, grams, and number of particles:

1. 9.35g of Al₂O₃ contains how many molecules?
2. 0.253 moles of BN has a mass of…
3. Calculate the number of moles of 8.46 x 10²⁴ atoms of F.
4. Calculate the number of moles of 4.35g of Cs₂CO₃.
5. 0.692moles of CsH contains how many molecules?
6. How many moles are in 14.49 g of CoSO₄.
7. What is the mass of 7.20 x 10²³ atoms of Ne?
8. 50.82 g of Cr₂O₃ has how many moles?
9. 2.84 moles of N₂H₄ has how many molecules?
10. 5.29 x 10²⁴ molecules of HCl has a mass of…

B. % Composition when given mass

1. In a 30.0 g sample of Pb(NO₃)_2, Pb has a mass of 2.535 g, N a mass of 8.694 g, and O a mass of 18.768 g. What is the % composition?
2. In a 25.0 g sample of Hg₂Cl₂, Hg has a mass of 3.755 g and Cl has a mass of 21.245 g. What is the % composition.
3. In a 20.0 g sample of Ni(NO₃)₂, Ni has a mass of 3.066 g, Ni has a mass of 6.426 g, and O has a mass of 10.508 g. What is the % composition?

C. % Composition when given empirical formula?

1. What is the % composition of KMnO₄?
2. What is the % composition of Sc₂O₃?
3. What is the % composition of H₂SeO₄?
4. What is the % composition of Cl₃HSi?
5. What is the % composition of NaBrO?

D. Finding empirical formula from % composition

1.  Na_Al_O_

• Na: 28.05%
• Al: 32. 92%
• O: 39.03%

2. Na_Mn_O_

• Na: 16.20%
• Mn: 38.71%
• O: 45.09%

3. S_O_Cl_

• S: 32.76%
• O: 23.71%
• Cl: 52.53%

4. Zr_O_

• Zr: 74.03%
• O: 25.97%

5. Pb_Cl_

• Pb: 75.51%
• Cl: 25.49%

6. H_Cl_O_

• H: 1.92%
• Cl: 67.58%
• O: 30.50%

7. B_H_

• B: 21.86%
• H: 78.14%

E. Stoichiometry 

1. Using the following equation, answer the questions:
    PCl₃ + 3H₂O ➝ H₃PO₃ + 2HCl

    A. How many moles of HCl are produced if 3.4 moles of H₂O are used?

    B. How many grams of PCl₃ are needed to produce 12.0 moles of HCl?

    C. How many grams of H₃PO₃ are produced if 13.40 g of H₂O are used?

2. Using the following equation, answer the questions:
    2CaH₂ + 5O₂ ➝ 4CO₂ + 2H₂O

    A. How many moles of O₂ are used to make 23.0 grams of H₂O?

    B. How many moles of CaH₂ must be used to make 12.0 moles of CO₂?

    C. How many grams of H₂O are made when 25.0 grams of CaH₂ are used?

ANSWERS:

A. Find moles, grams, and number of particles:

1. 5.52 x 10²² molecules of Al₂O₃
2. 6.28 g of BN
3. 14.05 mol F
4. 0.013 mol CsCO₃
5. 4.17 x 10²³ molecules CsH
6. 0.093 mol CoSO₄
7. 24.13 g Ne
8. 0.334 mol Cr₂O₃
9. 1.71 x 10²⁴ molecules N₂H₄
10. 320.28 g HCl

B. % Composition when given mass

1. Pb: 8.46%; N: 28.98%; O: 62.56% 
2. Hg: 15.02%; 84.98%
3. N: 15.33%; 32.13%; O: 52.54%

C. % Composition when given empirical formula?

1. K: 24.74%; Mn: 34.76%; O: 40.50%
2. Sc: 65.20%; O: 34.80%
3. H: 1.39%; Se: 54.47%; O: 44.14%
4. Cl: 78.52%; H: 0.74%; Si: 20.73%
5. Na: 19.34%; Br: 67.21%; O: 13.46%

D. Finding empirical formula from % composition

1. NaAlO₂
2. NaMnO₄
3. SO₂Cl₂
4. ZrO₂
5. PbCl₂
6. HClO
7. B₂H₆

E. Soichiometry

1A. 2.27 mol HCl

1B. 823.92 g PCl₃

1C. 20.34 g H₃PO₃

2A. 3.19 mol O₂

2B. 6.0 mol CaH₂

2C. 10.70 g H₂O

Practice 9.1

·         How many atoms are in 3.2 mols of Na?

·         How many atoms are in 1.9 mols of Au?

·         Calculate the mass of 0.2345 mols of fluorine atoms.

·         Calculate the mass of 1.575 mols of chromium atoms.

·         How many atoms are in 33.3 g of Fe?

·         How many atoms are in 2.9 g of N?

·         What is the molar mass of H₂SO₄?

·         What is the molar mass of KMnO₄?

Compounds and the Mole

·         If a substance has more than one element, add up the molar mass of the element.

·         For example: H₂O has two atoms of hydrogen and one atom of oxygen. You figure it out by adding the molar masses of the two different elements.

o   2H + O = molar mas of H₂O

o   2(1.01 g) + 16.00 g = 18.02 g/mol

·         These units are expressed in grams per mol because the molecular weight is per 1 mol of the substance.

·         Example Problem: Find the molar mass of Al₂(SO₄)₃.

o   First, we make our equation:

§  2Al + 3S + 12O = molar mass of Al₂(SO₄)₃

o   Second, we plug in our numbers:

§  2(26.98 g) + 3(32.07 g) + 12(16.00 g) = 342.2 g/mol

Molar Mass

• Compare the weight of 12 apples and 12 grapes. The apples would weigh considerable more. It is the same for elements. A mole of carbon is considerable lighter than one mole of copper.
• Scientists use the mole to relate to the atomic weight on the periodic table. If we say that Hydrogen has an atomic weight of 1.01 g, what does this mean? How much Hydrogen?
• The mass of one mole of any pure substance is called its molar mass.
• For example:
• Carbon has an atomic weight of 12.01 g. This means that 1 mole of Carbon atoms (or 6.022 x 10²³ atoms of Carbon) have a mass of 12.01 g.
• Mercury has an atomic weight of 200.59 g. This means that 1 mole of Mercury has a mass of 200.59 g.  
• Example problem: Calculate the mass of 0.5000 mol of Helium.
• 0.5000 mol He x 4.003 g He/1 mol He = 2.002 g
• Example problem: How many copper atoms are in a 4.00 g sample of pure copper wire.
• First, we need to find the number of moles.
• 4.00 g Cu x 1 mol Cu/63.55 g Cu = 0.0629 mol Cu
• Second, using Avogadro’s number, we can find the number of atoms.
• 0.0629 mol Cu x 6.0022 x 10²³ atoms Cu/1 mol Cu = 3.79 x 12²² atoms Cu

Avogadro’s Number and the Mole

• Because atoms are so small and nearly impossible to count, scientists have developed a unit called the mole to make counting atoms more practical.
• A mole (mol) is the amount of substance contained in 6.022 x 10²³ particles. This is called Avogadro’s number.
• 1 mol of He atoms = 6.022 x 10²³ He atoms
• 1 mol of H₂O molecules = 60.22 x 10²³ H₂O molecules
• 1 mole of NaCl formula units = 6.022 x 10²³ NaCl formula units
• Because the mole contains such a huge number of particles, it is only used to describe things that are very small.
• Example Problem: How many atoms are in a 4.5 mol sample of helium? 2.7 x 10²⁴ atoms

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