Quantitative Composition of Compounds

Making new chemicals is much like following a recipe from a cook book...1 cup flour + 2 eggs + ½ tsp baking powder 5 pancakes

What if you want to make more (or less)?Suppose you have plenty of flour and baking powder, but only 8 eggs. How many pancakes can you make?

Solve it in your head: 2 eggs makes 5 pancakes, so four times more eggs

You can solve it using conversion factor:5 pancakes 2 eggs

makes 20 (5x4) pancakes.

8 eggs5 pancakes 2 eggs

x = 20 pancakes

Chapter 7

… except you don’t get to lick the spoon!

3 blocks cream cheese + 5 eggs + 1 cup sugar = 1 cheese cake.

How many cheese cakes can we make out of 15 eggs?

How much sugar do we need for 5 cheese cakes? (5)

Practice using the following mouth-washing, diet-buster recipe:

15 eggs x 1 cake5 eggs = 3 cheese cakes

Suppose you want to ‘whip’ a batch of hydrogen iodide, following the balanced chemical equation: H2 + I2 2 HI

How much H2 and I2 should you use to make 10 g of HI?

A common mistake is that H2 and I2 react in one-to-one mass ratio so:

5 g H2 + 5 g I2 10 g HIThe coefficients balancing the equation refer to number of atoms, not masses.

Introducing the mole. The mole is like a dozen, but much, much more.

1 mole of anything:

donuts, pancakes, atoms, molecules, ions…

is always 6.022 x 1023 of that thing.

1 mole of soft drink cans is enough to cover the surface of the earth to a depth of over 200 miles.

If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

The mole is Avogadro’s Number of items. 1 mole = 602,214,179,000,000,000,000,000 or 6.022 x 1023.We need the mole because the mass of an atom is too small to be measured on a balance. Remember: 1 amu = 1.6 x 10-24 g.

6.022 x 1023 6.022 x 1023 1.204 x 1024

molecules molecules molecules

H2 + I2 = 2 HI

The mole translates between the number of atoms (or molecules, ions) and grams of atoms (molecules, ions).

It is defined as the mass of Avogadro’s number of atoms 6C, which, in turn, weights exactly 12 g.12

A mole of atoms weighs the same number of grams as the atomic mass.

One mole of H atoms weighs 1.0079 g.One mole of C atoms weighs 12.011 g.

Atomic mass refers to: the sum of protons and neutrons in a single atom, weighted average mass of all isotopes of an element and also to the number of grams in one mole of atoms.

1 molecule 1 molecule 2 molecule2 H atoms 2 I atoms 2 x (1 atom H, 1 atom I)

Conversion factors:

1 mole6.022 x 1023 species

1 molemolar mass1 mole of H2 weighs 2 x 1.0079 g = 2.0158 g

Amadeo Avogadro The mole

or any number of molecules

12 molecules 12 molecules 24 molecules

1 mole 1 mole 2 mole 2.0158 g 253.81 g 255.8258 g

Mole - mass - atoms Q1: How many atoms in 0.5 mole Au?

0.5 mole Au x6.022 x 1023 atoms Au 1 mole Au

= 3.011 x 1023 atoms Au

Q2: What is the mass of 0.5 mol Au?

0.5 mole Au x196.967 g Au 1 mole Au

= 98.4835 g Au

Q3: How many atoms in 15.00 g Au?

15.00 g Au x 1 mole Au196.987 g Au

6.022 x 1023 atoms Au 1 mole Aux = 4.59 x 1022 atoms Au

Q1a: How many moles in 7.12 x 1024 atoms of Cu?

7.12 x 1024 atoms Cu x 1 mole Cu 6.022 x 1023 atoms Cu

= 11.8 mol Cu

conversions

Percent Composition What is the % composition of CH2O?

Total mass = 12.01 g + 2.016 g + 16.00 gPercent composition is % mass that each element in a molecule contributes to the total molar mass of the compound. Assume that you have one mole of the compound.

%C = 12.01 g30.026 g

x 100

%H = 6.71 %

%O = 53.29 %

100.00 %

+Practice: What is the % composition of glucose? Check your answer: it is the same as in CH2O!

Empirical Formula: the formula of a compound that expresses the smallest whole number ratio of the atoms present.

Types of Formulas

Molecular Formula: the formula that states the actual number of each kind of atom found in one molecule of the compound.

CH2O is the empirical formula for glucose, C6H12O6

Formulas describe the relative number of atoms (or moles) of each element in a formula unit. It’s always a whole number ratio.

1 molecule of C9H8O4 = 9 atoms of C, 8 atoms of H and 4 atoms of O.

1 mole of C9H8O4 = 9 mol of C, 8 mol of H and 4 mol of O atoms.

If we can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

= 30.026 g

%C = 40.00 %

Dr. Ent burned 0.5 g of the sample and obtained the total of over 1 g of products. How is that possible?

From the mass of the products (water and carbon dioxide) we determine the number of moles of C, H, and O, and from them obtain the empirical formula of the compound.

Oxygen from air is a reactant!

1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”.

2. Convert grams of CO2 and H2O (or C and H) into moles of C and H atoms.3. Convert moles of C into grams of C. Do the same for H.

6. Divide the molar mass of the compound by the molar mass of the empirical formula. To find the molecular formula, multiply all subscripts in the empirical formula by this product.

Combustion analysis shown 0.300 g H2O and 0.733 g CO2 from 0.500 g of sample. Find the empirical and molecular formula if the molar mass of the compound is 180.15 g/mol.

2.

0.3 g H2O x 1 mol H2O18.01 g H2O

x2 mol H atms1 mol H2O

= 0.0333 mol H at.

0.733 g CO2 x 1 mol CO2

44.01 g CO2

x1 mol C atms1 mol CO2

= 0.0166 mol C at.

3.

0.0333 mol H x1.008 g H 1 mol H = 0.0336 g H

0.0166 mol C x12.01 g C 1 mol C = 0.199 g C

g O = 0.5 – (0.0336 + 0.199) = 0.267 g O

4. Add masses for C and H and subtract the sum from the mass of the sample to obtain mass of O. Convert the mass into moles of O.

4.0.267 g O x

1 mol O at.16.00 g O = 0.0169 mol O at.

5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula.

5.H: 0.0333 / 0.0166 = 2C: 0.0166 / 0.0166 = 1O: 0.0169 / 0.0166 ~ 1 CH2O

Empirical formula

6.

Molar massemp. form.

30.026

Molar mass sampleMolar mass emp. formula

= 180.15 30.03

= 6

Molecular formula: C6H12O6.

Note: steps 3, 4 apply only for finding formulas from combustion analysis. There is 6 CH2O units

in the compound.

3. Percent composition of a compound is found to be 43.2% K, 39.1% Cl, and some O. Find the empirical formula. If the molar mass of the compound is 90.550 g mol-1, find the molecular formula. (KClO)

Find the empirical and molecular formulas if the % composition is 40.0% C, 6.70% H, 53.3% O, and the molar mass of the compound is 180.155 g/mol.

2. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34 g O. Determine an empirical formula for this substance. (NO2)

1. A compound has an empirical formula of NO2. The colorless liquid used in rocket engines has a molar mass of 92.0 g mole-1. What is the molecular formula of this substance? (N2O4)

1. Determine the mass in grams of each element present, if necessary. Remember, % means “out of 100”.2. Convert grams of CO2 and H2O (or C and H) into moles of C and H atoms.3. Convert moles of C into grams of C. Do the same for H.

6. Divide the molar mass of the compound by the molar mass of the empirical formula. To find the molecular formula, multiply all subscripts in the empirical formula by this product.

4. Add masses for C and H and subtract the sum from the mass of the sample to obtain mass of O. Convert the mass into moles of O.

5. Divide all number of moles with the smallest to obtain the subscripts of the empirical formula.

1. Assume that you have 100.00 g sample; the mass of each element is equal to the % composition. 40.0 g C, 6.70 g H, 53.3 g O.

2.

40.0 g C x 1 mol C12.01 g C 6.70 g H x

1 mol H1.008 g H 53.3 g O x

1 mol O16.00 g O

= 3.33 mol C = 6.65 mol H = 3.33 mol O

Skip steps 3 and 4, they apply for combustion analysis only.

5. C: 3.33 / 3.33 = 1H: 6.65 / 3.33 = 2O: 3.33 / 3.33 = 1

Empirical formula CH2O.

Emp. Formula mass = 30.026

6. Molar mass sampleMolar mass emp. formula

= 180.155 30.026

= 6

Molecular formula: C6H12O6.

Practice (answer in parenthesis):

Than

k you

, Dr.

Ent!

Thus, there are 6 (CH2O) units.

Chapter 9 Calculations from Chemical Equations

Remember me?

The molar mass of an element is its atomic mass in grams.It contains 6.022 x 1023 atoms (Avogadro’s number) of the element.The molar mass of a compound is the sum of the atomic masses of all its atoms.

Conversions go through moles.For calculations of mole-mass-number_of_particle relationships:

1. Use balanced equation.

2. The coefficient in front of a formula represents the number of moles of the reactant or product.

Al + Fe2O3 Al2O3 + Fe

2 2

2 mol 1 mol 1 mol 2 mol

To quantitatively convert from one quantity to another we introduce mole ratio:

Mole ratio is found from the coefficients of the balanced equation.

1 mol Fe2O3

2 mol Al1 mol Fe2O3

1 mol Al2O3

Mole ratio = moles of desired substancemoles of starting substance

A mole of a compound weighs the sum of all atoms in the compound.

For instance: molar mass of NaCl is 22.99 + 35.45 = 65.44 g

Which conversion factor will be used depends on starting and desired substance

Mole – Mole Conversions - MoleculesExample 1:How many moles of NaCl result from the complete reaction of 3.4 mol of Cl2? Assume that there is more than enough Na. 2 Na(s) + Cl2(g) 2 NaCl(s)

1 mole 2 moles

3.4 moles Cl2 x2 moles NaCl 1 mole Cl2

= 6.8 moles NaCl

Ca5(PO4)3F(s) + 5H2SO4(aq) 3H3PO4(aq) + HF(aq) + 5CaSO4(s)

Example 2: Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4) on phosphate rock:

1 mole 5 moles 3 moles 1 mole 5 moles

3 moles H3PO4

5 moles H2SO4

10 moles H2SO4 x = 6 moles H3PO4

Example 3: Calculate the number of moles of Ca5(PO4)3F needed to produce 6 moles of H3PO4.

1 mole Ca5(PO4)3F 3 moles H3PO4

6 moles H3PO4 x = 2 moles Ca5(PO4)3F

desired substance

starting substance

The following examples refer to the equation:

Example 4: Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4.

Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4

1 mole 5 moles 3 moles 1 mole 5 moles

1. Convert the starting substance into moles. 2. Convert moles of starting substance into moles of desired substance.3. Convert moles of desired substance into the units specified in the problem.

Molar mass of H3PO4 = 97.994

784 g H3PO4 x 1 mole H3PO4

97.994 g H3PO4

= 8.00 moles H3PO4.

gmole

8.00 moles H3PO4 x 5 moles H2SO4

3 moles H3PO4

= 13.3 moles H2SO4.

done.

Mass – Mole conversion

Ex. 5: Calculate the mass of phosphate rock, Ca5(PO4)3F needed to yield 200. g of HF.Molar masses: Ca5(PO4)3F = 504.31 g/mol; HF = 20.008 g/mol

Step 1, 200. g HF x 1 mole HF20.008 g HF

= 10.0 moles HF xStep 2

1 mole Ca5(PO4)3F 1 mole HF

= 10.0 molesCa5(PO4)3F

Step 3:10.0 moles Ca5(PO4)3F x

504.3 g ph.r.1 mole ph.r

= 5.00 kg Ca5(PO4)3F.

Mass – mass conversionEx. 6: Calculate the number of grams of H2SO4 necessary to yield 392 g of H3PO4.

Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4

1 mole 5 moles 3 moles 1 mole 5 moles

1. Convert the starting substance into moles.

2. Convert moles of starting substance into moles of desired substance.

3. Convert moles of desired substance into the units specified in the problem.

Molar mass H3PO4 = 97.994

392 g H3PO4 x 1 mole H3PO4

97.994 g H3PO4

= 4.00 moles

gmole

4.00 moles H3PO4 x 5 moles H2SO4

3 moles H3PO4= 6.67 moles

Molar mass H2SO4 = 98.086 gmole

6.67 moles H2SO4 x 98.086 g1 mole H2SO4

= 654 g H2SO4.

Example 7: Find the mass of glucose that can be synthesized from 58.5 g of CO2, assuming that there is more than enough water to react with all the CO2. Molar masses are 44.01 g (CO2) and 180.16 (glucose).

(aq)(g)(l)(g) 61262sunlight

22 OHC O 6 OH 6 CO 6

58.5 g CO2 x 1 mole CO2

44.01 g CO2

x1 mole glucose 6 moles CO2

180.16 g glucose 1 mole glucose

x = 39.9 g glucose

392 g H3PO4 x 1 mole H3PO4

97.994 g H3PO4

5 moles H2SO4

3 moles H3PO4

98.086 g1 mole H2SO4

= 654 g H2SO4.

Step_by_step:

Combined steps:

x x

Conversion – General Case

Example 8: Calculate the mass of NH3 formed by the reaction of 112 grams of H2. N2 + 3H2 2NH3 grams H2 moles H2 moles NH3 grams NH3

Mass to molesof starting compound

Moles of starting compoundto moles of desired compound

Moles of desired comp.to units desired.

Step 1 Step 2 Step 3

Moles – moles: Step 2 only

Moles – mass: Step 2 and Step 3 only

Mass – mass: All 3 steps

Molar masses: H2: 2.016 g/mol; NH3: 17.034 g/mol

112 g H2 x 1 mole H2

2.016 g H2

Step 1

2 moles NH3

3 moles H2

x

Step 2

17.034 g NH3

1 mole NH3

x

Step 3

= 1420 g NH3 = 1.42 kg NH3.

Example 9: Calculate the moles of NH3 formed by the reaction of 1.5 moles of H2.

2 moles NH3

3 moles H2

x

Step 2

= 17.0 g NH3.1.50 moles of H2

Startingcompound

Example 10: Calculate the mass of NH3 formed by the reaction of 1.50 moles of H2.

Startingcompound

2 moles NH3

3 moles H2

x

Step 2

= 1.00 mole NH3.1.50 moles of H2

Startingcompound

17.034 g NH3

1 mole NH3

x

Step 3

result

result

result

Mass – moles: Step 1 and Step 2 only

2 moles NH3

3 moles H2

x

Step 2

= 49.6 g NH3.150. g H2

Example 11: Calculate the moles of NH3 formed by the reaction of 150. g H2.

Startingcompound

x

Step 1 result

Conversion – General Case (cont’d)

N2 + 3H2 2NH31 mole H2

2.016 g H2

Mass – particles: All 3 steps

112 g H2 x 1 mole H2

2.016 g H2

Step 1

2 moles NH3

3 moles H2

x

Step 2

6.022 x 1023 molecules NH3

1 mole NH3

x

Step 3

= 2.23 x 1025 molecules NH3.Starting

compound result

Example 12: Calculate the # molecules of NH3 formed by the reaction of 150. g H2.

Limiting Reactant and Yield Calculations

The amount of the product(s) depends on the reactant that is used up during the reaction, i.e. limiting reactant.

The number of seats is the limiting part (reactant); one frame and two wheels are parts in excess; 3 bicycles is the yield.

One bicycle needs 1 frame, 1 seat and 2 wheels, therefore not more than 3 bicycles can be made.

Example 13: How many moles of Fe3O4 can be obtained by reacting 16.8 g Fe with 10.0 g H2O? Which substance is the limiting reactant? Which substance is in excess? How much of the reactant in excess remains unreacted?

Limiting Reactant and yield Calculations (cont’d)

Balanced equation: 3 Fe (s) + 4 H2O (g) Fe3O4 (s) + 4 H2 (g)

16.8 g Fe x 1 mol Fe55.85 g Fe

x 1 mol Fe3O4

3 mol Fe = 0.100 mol Fe3O4.

Strategy: 1. Write and balance equation.2. Calculate the number of moles of product for each reactant;3. The reactant that gives the least moles of (the same!) product is the limiting reactant. 4. Find the amount of reactant in excess needed to react with the limiting reactant.

Subtract this amount from the starting quantity to obtain the amount in excess.5. Find the yield from the limiting reactant.

from Fe:

10.0 g H2O x 1 mol H2O18.02 g H2O

x 1 mol Fe3O4

4 mol H2O= 0.139 mol Fe3O4. From H2O:

limiting reactant

yield

Yield is 0.100 mol Fe3O4, Fe is the limiting reactant, 2.99 g H2O is in excess.Answer:

16.8 g Fe x 1 mol Fe55.85 g Fe

x 4 mol H2O 3 mol Fe

x18.02 g H2O 1 mol H2O

= 7.01 g H2O.ReactedH2O

10.0 g – 7.01 g = 2.99 g H2O.Excess:

Least moles Fe3O4?

Yield

Many reactions (especially organic) do not give the 100% yield, due to:side reactions, reversible reactions, product losses due to human factor.

Percent Yield

Calculations done so far assumed that the reaction gives maximum (100%) yield.

Theoretical yield: Amount calculated from the chemical equation.

Actual yield: Amount obtained experimentally.

Percent yield:Actual yieldTheor. yield x 100 %

Example 14: If 65.0 g CCl4 was prepared by reacting 100. g CS2 and 100. g of Cl2, calculate the percent yield.

CS2 + 3 Cl2 CCl4 + S2Cl2

Molar masses: CS2: 76.15; Cl2: 70.90; CCl4: 153.81 g/mol

100. g CS2 x1 mol CS2

76.15 g CS2

x1 mol CCl41 mol CS2

= 1.31 mol CCl4.

100. g Cl2 x1 mol Cl270.90 g Cl2

x1 mol CCl4 3 mol CS2

= 0.470 mol CCl4.

Theoretical yield

Limiting reactant

0.470 mol CCl4 x153.81 g CCl4 1 mol CCl4

= 72.3 g CCl4.

65.0 g CCl472.3 g CCl4

x 100 % = 89.9 %Percent yield

Strategy:Find limiting reactant.Calculate theoretical yield.Calculate percent yield.

65.0 g CCl4Actual yield

HW, Chp. 7: 1, 5, 15, 26, 33 Chp. 9: 3, 7, 13, 15, 23, 29