stoichiometry © 2009, prentice-hall, inc. warmup – mass percent find the percent by mass: 1.what...

Stoichiometry

© 2009, Prentice-Hall, Inc.

Warmup – Mass Percent

Find the percent by mass:

1. What is the mass % of silver in silver nitrate? (63.5%)

2. What is the mass % of N in ammonium carbonate? (29.2%)

Stoichiometry


Finding Empirical Formulas

Stoichiometry


Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition.

Stoichiometry


Rhyme

Rhyme to remember order of steps to convert % composition empirical formula:

• Percent to mass

• Mass to mole

• Divide by small

• Times till whole

Stoichiometry


Sample Exercise 3.13 (p. 99)

Ascorbic acid (vitamin C) contains 40.92% C, 4.58% H, and 54.50% O by mass.

What is the empirical formula of vitamin C?(C3H4O3)

Stoichiometry



Assuming 100.00 g of ascorbic acid,

C: 40.92 g x = 3.407 mol C

H: 4.58 g x = 4.53 mol H

O: 54.50 g x = 3.406 mol O

1 mol12.01 g

1 mol16.00 g

1 mol1.01 g

Stoichiometry


Calculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:

C: = 1.000

H: = 1.33

O: = 1.000

Multiply all by 3: 3 C:4 H:3 O

3.407 mol3.406 mol

4.53 mol3.406 mol

3.406 mol3.406 mol

Stoichiometry



These are the subscripts for the empirical formula:

C3H4O3

Stoichiometry


Practice Exercise 1 (3.13)A 2.144-g sample of phosgene, a compound used as a chemical warfare agent during World War I, contains 0.260 g of carbon, 0.347 g of oxgen, and 1.537 g of chlorine. What is the empirical formula of this substance?

(a)CO2Cl6,

(b)COCl2,

(c)C0.022O0.022Cl0.044,

(d)C2OCl2

Stoichiometry


Practice Exercise 2 (3.13)

A 5.325-g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?

(C4H4O)

Stoichiometry


Molecular Formula from Empirical Formula

The empirical formula (relative ratio of elements in the molecule) may not be the molecular formula (actual ratio of elements in the molecule).

e.g. ascorbic acid (vitamin C) has empirical formula C3H4O3.

The molecular formula is C6H8O6.

Stoichiometry


Molecular Formula from Empirical Formula

To get the molecular formula from the empirical formula, we need to know the molecular weight, MW.

(“molecular molar mass”)

The ratio of molecular weight (MW) to formula weight (FW) of the empirical formula must be a whole number.

(“formula molar mass”)

Stoichiometry



Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4. The experimentally determined molecular weight of this substance is 121 amu.

What is the molecular formula of mesitylene?

(Hint: you do not need to figure out the empirical formula first)

(C9H12)

Stoichiometry



Cyclohexane, a commonly used organic solvent, is 85.6% C and 14.4% H by mass with a molar mass of 84.2 g/mol. What is its molecular formula?

(a)C6H,

(b)CH2,

(c)C5H24,

(d)C6H12,

(e)C4H8

Stoichiometry



Ethylene glycol, the substance used in automobile antifreeze, is composed of 38.7% C, 9.7% H, and 51.6% O by mass. Its (molecular) molar mass is 62.1 g/mol.

a) What is the empirical formula of ethylene glycol? (CH3O)

b) What is its molecular formula? (C2H6O2)

Stoichiometry


Hydrates

• Anhydrous = dry, does not contain water

• Hydrate = a compound formed by the combination of water with some other substance in which the water retains its molecular state as H2O

Stoichiometry


Determining Hydrate Formulas

1. Find the formula of a hydrate from experimental mass data:a) Calculate the number of moles of the anhydrous

salt (mass mole conversion)

b) Calculate the mass of H2O by subtracting mass of anhydrous salt from mass of hydrated salt

c) Calculate the moles of H2O (mass mole conversion)

d) Calculate ratio: # mol H2O:# mol anhydrous salt

e) correct formula = salt · x H2O (x = mole ratio)

Stoichiometry



2. Find the formula from mass % data:a) Use the method of using 100 g (from 100 %) to

give mass in grams of the salt and waterUse mass mole conversion to calculate the # moles of the anhydrous salt and the # moles

of water.

b) Calculate the mole ratio of H2O:anhydrous salt

Stoichiometry



3. Find the % H2O in a hydrate from

its formula: Use mass percent.

Stoichiometry


Hydrate practice problems

1. A mass of 2.50 g of blue hydrated copper sulfate (CuSO4 • x H2O) is placed in a crucible and heated. After heating, 1.59 g white anhydrous copper sulfate (CuSO4) remains. What is the formula for the hydrate? Name the hydrate.

(CuSO4 • 5 H2O)

Stoichiometry



2. A hydrate is found to have the following percent composition:

48.8% MgSO4 and 51.2% H2O.

What is the formula and the name for this hydrate?

(MgSO4 • 7 H2O)

Stoichiometry



3. The formula for a hydrate is SnCl2 • 2 H2O.

What is the mass percent of water in this hydrate?

(15.97% H2O)

Stoichiometry


Combustion Analysis

• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this.– C is determined from the mass of CO2 produced.

– H is determined from the mass of H2O produced.

– O is determined by difference after the C and H have been determined.

Stoichiometry



Isopropyl alcohol, a substance sold as rubbing alcohol, is composed of C, H, and O. Combustion of 0.255 g of isopropyl alcohol produces 0.561 g CO2 and 0.306 g H2O. Determine the empirical formula of isopropyl alcohol.

(C3H8O)

Stoichiometry



The compound dioxane, which is used as a solvent in various industrial processes, is composed of C, H, and O atoms. Combustion of a 2.2.03-g sample of this compound produces 4.401 g CO2 and 1.802 g H2O. A separate experiment shows that it has a molar mass of 88.1 g g/mol. Which of the following is the correct molecular formula for dioxane?

(a)C2H4O,

(b)C4H2O2,

(c)CH2,

(d)C4H8O2

Stoichiometry



a) Caproic acid, which is responsible for the foul odor of dirty socks, is composed of C, H, and O atoms. Combustion of a 0.225 g sample of this compound produces 0.512 g CO2 and 0.209 g H2O. What is the empirical formula of caproic acid?

(C3H6O)

b) Caproic acid has a molar mass of 116 g/mol. What is its molecular formula?

(C6H12O2)

Stoichiometry


Elemental Analyses

Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

Stoichiometry


Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometry



Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).

Stoichiometry



Starting with 1.00 g of C6H12O6… we calculate the moles of C6H12O6…use the coefficients to find the moles of H2O…and then turn the moles of water to grams.

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Stoichiometry


Limiting Reactants

Stoichiometry


How Many Cookies Can I Make?

• You can make cookies until you run out of one of the ingredients.

• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).

Stoichiometry


How Many Cookies Can I Make?

• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.

Stoichiometry

Stoichiometry Balloon Races


Stoichiometry


Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount.– In other words, it’s the reactant you’ll run out of first (in

this case, the H2).

Stoichiometry


Limiting Reactants

In the example below, the O2 would be the excess reagent.

Stoichiometry


Solving Limiting Reactant Problems

Given: 8.51 g H2 and 9.25 g O2

2 H2 + O2 2 H2O

Which reactant is limiting?

How much excess is left over?

Stoichiometry


One Method


2 H2 + O2 2 H2O

g A have g B need Is this more or less than what you have?

8.51 g H2 (1 mol H2)(1 mol O2)(32.00 g O2) = 67.4 g O2

HAVE 2.02 g H2 2 mol H2 1 mol O2 NEED 67.4 g O2 > 9.25 g O2

NEED > HAVE

O2 is limiting

Stoichiometry



2 H2 + O2 2 H2O

Because O2 is limiting, Step 1

9.25 g O2 (1 mol O2)(2 mol H2)(2.02 g H2) = 1.17 g H2 used 32.00 g O2 1 mol O2 1 mol H2

Step 2 8.51 g H2 HAVE-1.17 g H2 USED 7.34 g H2 left over = excess reactant

Stoichiometry


We can also use a tool called a RICE table

R = reactionI = initial quantitiesC = changeE = end quantities (later in the year = equilibrium

Stoichiometry


R 2 H2 + O2 2 H2O

I 4.21 mol 0.289 mol 0

C - 0.578 mol -0.289 mol +0.578 mol

E 3.63 mol 0 mol 0.578 mol limiting theoretical reactant yield

Take coefficientsinto account in thisstep

Supporting calculations:

8.51 g H2(1 mol H2) = 4.21 mol H2

2.02 g H2

9.25 g O2(1 mol O2) = 0.289 mol O2

32.00 g O2

Resulting calculations:

3.63 mol H2(2.02 g H2) = 7.34 g H2 1 mol H2 in excess

0.578 mol H2O(18.02 g H2O) = 10.4 g H2O product

Stoichiometry



The most important commercial process for converting N2 from the air into nitrogen-containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):

N2(g) + 3 H2(g) 2 NH3(g)

How many moles of NH3 can be formed from 3.0 mol of N2, and 6.0 mol of H2?

(4.0 mol NH3)

Stoichiometry


Practice Exercise 1 (3.18)When 24 mol of methanol and 15 mol of oxygen combine in the combustion reaction

2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g),

what is the excess reactant and how many moles of it remains at the end of the reaction?

(a)9 mol CH3OH(l),

(b)10 mol CO2(g),

(c)10 mol CH3OH(l),

(d)14 mol CH3OH(l),

(e)1 mol O2(g)

Stoichiometry



Consider the reaction 2 Al(s) + 3 Cl2(g) 2 AlCl3(s).

A mixture of 1.50 mol of Al and 3.00 mol of Cl2 are allowed to react.a) What is the limiting reactant? (Al)b) How many moles of AlCl3 are formed? (1.50 mol)c) How many moles of the excess reactant

remain at the end of the reaction? (0.75 mol Cl2)

Stoichiometry



Consider the following reaction that occurs in a fuel cell:

2 H2(g) + O2(g) 2 H2O(l)

This reaction, properly done, produces energy in the form of electricity and water. Suppose a fuel cell is set up with 150 g of hydrogen gas and 1500 grams of oxygen gas (each measurement is given with two significant figures). How many grams of water can be formed?

(1300 g H2O)

Stoichiometry


Practice Exercise 1 (3.19)Molten gallium reacts with arsenic to form the semiconductor, gallium arsenide, GaAs, used in light-emitting diodes and solar cells: Ga(l) + As(s) GaAs(s)

If 4.00 g of gallium is reacted with 5.50 g of arsenic, how many grams of the excess reactant are left at the end of the reaction?

(a)4.94 g As,

(b)0.56 g As,

(c)8.94 g Ga,

(d)1.20 g As.

Stoichiometry



A strip of zinc metal weighing 2.00 g is placed in an aqueous solution containing 2.50 g of silver nitrate, causing the following reaction to occur:

Zn(s) + 2 AgNO3(aq) 2 Ag(s) + Zn(NO3)2(aq)

a) Which reactant is limiting? (AgNO3)b) How many grams of Ag will form? (1.59 g)

c) How many grams of Zn(NO3)2 will form? (1.39 g)d) How many grams of the excess reactant will be

left at the end of the reaction? (1.52 g Zn)

Stoichiometry


Theoretical Yield

• The theoretical yield is the maximum amount of product that can be made.– In other words it’s the amount of product

possible as calculated through the stoichiometry problem.

• This is different from the actual yield, which is the amount one actually produces and measures.

Stoichiometry


Percent Yield

One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield).

Actual YieldTheoretical YieldPercent Yield = x 100

Stoichiometry



Adipic acid, H2C6H8O4, is used to produce nylon. It is made commercially by a controlled reaction between cyclohexane (C6H12) and O2:

2 C6H12(l) + 5 O2(g) 2 H2C6H8O4(l) + 2 H2O(g)

a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane, and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (43.5 g H2C6H8O4)

b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid? (77.0%)

Stoichiometry



If 3.00 g of titanium metal is reacted with 6.00 g of chlorine gas, to form 7.7 g of titanium(IV) chloride in a combination reaction, what is the percent yield of the product?

(a)65%,

(b)96%,

(c)48%,

(d)86%

Stoichiometry



Imagine that you are working on ways to improve the process by which iron ore containing Fe2O3 is converted into iron. In your tests you carry out the following reaction on a small scale:

Fe2O3(s) + 3 CO(g) 2 Fe(s) + 3 CO2(g)

a) If you start with 150 g of Fe2O3 as the limiting reagent (reactant), what is the theoretical yield of Fe? (105 g Fe)

b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? (83.7%)

stoichiometry © 2009, prentice-hall, inc. warmup – mass percent find the percent by mass: 1.what...

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