Chapter 4

Quantities of Reactants and Products

Antoine Lavoisier 1743-1794. “Father

of modern chemistry.” Recognized true elements.

Used quantitative measurements in chemical reactions.

1

Chemical Equations

• • •

Lavoisier: mass is conserved in a chemical reaction.

Chemical equations: descriptions of chemical reactions.

Two parts to an equation: reactants and products : 2H 2 + O 2 2H 2 O

Reactants Product

2

H 2 O 2 H 2 O 3

•

Stoichiometric coefficients : numbers in front of the chemical formulas give numbers of molecules or atoms reacting (and numbers being produced).

4

Law of Conservation of Mass:

All reactions must be balanced

CH 4 + O 2 CO is not balanced. (Why?) 2 + H 2 O Count atoms: Reactants: Products: 1 C 1 C 4 H 2 O 2 H 3 O 5

Balance reactions only by changing coefficients, not by altering chemical formula 6

Combustion is the burning of a substance in oxygen: C 3 H 8 (g) + 5O 2 (g)



3CO 2 (g) + 4H 2 O(l) 7

Which is correct? (Blue=A; Red=B) a) A 2 b) A 2 + B A + 4B c) 2A + B 4 d) A + B 2 2 B 2 AB 2 AB AB 2 2 2 8

Atomic and Molecular Weights

Percentage Composition from Formulas Percent composition is the atomic weight for each element divided by the formula weight of the compound multiplied by 100:

% Element

 

Atoms of FW of Element



AW

 

Compound 100

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Percentage Composition from Formulas

What is % O in H 2 SO 4 (by mass)?

FW= (2x1) + (1x32)+ (4x16)= 98 amu FW of O in H 2 SO 4 = 4 x 16 = 64 amu %O = 64 x 100 = 65.3% 98 10

The Mole*

*MSJ Ch 3 pp 100-104

•

The “amu” is an “atomic mass unit.”

•

O has a mass of 16 amu – but we can’t weigh out anything in amu

•

If we want to keep the number “16” for the mass of oxygen in some real units (like grams) then we are dealing with a whole bunch of atoms (in 16 g of oxygen).

•

That bunch of atoms is called a mole .

•

Experimentally, 1 mole = 6.02 x 10 23 things (atoms)

•

This number is called Avogadro’s number.

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The mole is defined so that one mole of a substance has a mass equal to its AW or MW in grams

Basically, you are replacing amu with grams, e.g.

The mass of a P atom is 31 amu.

The molar mass of P is 31 grams. The mass of a Ca atom is 40 amu The molar mass of Ca is 40 grams This amount of P or Ca each contains Avogadro’s number (6.02 x 10 23 ) of atoms of P and Ca.

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The Mole

Experimentally, 1 mole of 12 C has a mass of exactly 12 g. (recall from Ch. 2) Molar Mass Molar mass: mass in grams of 1 mole of substance Units: g/mol or g.mol

-1 .

Mass of 1 mole of 12 C = 12 g exactly 13

14

This photograph shows one mole of : solid

NaCl

(58.5 g), liquid

H 2 O

(18 g), and gaseous

N 2

(28 g).

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The Mole

Molar Mass Molar mass: sum of the molar masses of the atoms: What is Molar Mass of H 2 SO 4 ?

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The Mole

Interconverting Masses, Moles, and Numbers of Particles 17

Example 1: 5.00 g of P (a) contains (b) contains mol of P atoms of P Example 2: 5.00 x 10 24 atoms of C (a) equals mol of C (b) has mass equal to grams.

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Example 3: 3.5 mol CO 2 : (a) has what mass?

(b) contains how many molecules of CO 2 ?

(c) contains how many atoms of O?

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Empirical Formulas from Analyses

Start with mass % of elements (i.e. empirical data) and calculate a formula.

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Empirical Formulas from Analyses*

Example: compound of N and O Given analysis: N: 25.9%; O: 74.1% Assume 100g; N: 25.9 g; O: 74.1 g Change to mol: N: 25.9 g x mol = 1.85 mol 14 g

*MSJ Ch 3 pp 104-108

O: 74.1 g x mol = 4.63 mol 16 g Preliminary emp. Formula: N 1.85

O 4.63

Clean it up: divide both by 1.85: Get N 1 O 2.5

; get rid of fractions, multiply both by 2: Get N 2 O 5 which is the empirical (simplest) formula

What are some possible molecular formulas?

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Empirical Formulas from Analyses

Molecular Formula from Empirical Formula Once we know the empirical formula, we need the MW to find the molecular formula.

Subscripts in the molecular formula are always whole number multiples of subscripts in the empirical formula.

Example: suppose compound of C and H has empirical formula of C 3 H 8 and a MW = 176 g/mol. What is the molecular formula?

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Quantitative Information from Balanced Equations

Balanced chemical equation gives number of molecules (or moles) that react to form products.

Interpretation: balanced equation gives us the ratio of number of moles of reactant to product (or v.v.) .These ratios are called stoichiometric ratios . Example: 2 H 2 + O 2 2 H 2 O Molecules: 2 1 2 Moles: 2 1 2 Ratio of O 2 :H 2 O = 1:2 (either molecules or moles) 23

Quantitative Information from Balanced Equations

The ratio of grams of reactant cannot be directly related to the grams of product.

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Stoichiometry

Problem: aluminum sulfide + water aluminum hydroxide + hydrogen sulfide (a) Write balanced reaction: (b) How many g aluminum hydroxide obtained from 10.5 g of aluminum sulfide?

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Stoichiometry

Problem: 2 NaN 3 (s) 2Na(s) + 3 N 2 (g) (a) how many mol N 2 produced from 2.50 mol NaN 3?

(b) how many g NaN 3 needed to form 6.00 g N 2 (c) how many g NaN 3 (1.00 ft 3 needed to produce 10.0 ft = 28.3 L; density of N 2 = 1.25 g/L) 3 of N 2 ?

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Limiting Reactants

If the reactants are not present in stoichiometric amounts, at end of reaction some reactants are still present (in excess).

Limiting Reactant : one reactant that is consumed.

H 2 O 2 O 2 INXS 27

RECIPE: 3 cups flour + 4 eggs + 2 cups sugar cake ------------------------------------------------------- On Hand: cups flour 20 eggs 20 cups sugar 20 How many cakes can be made from these amounts?

What is the “limiting reactant” (LR) What’s left over and how much of it is left over?

28

RECIPE: 3 cups flour + 4 eggs + 2 cups sugar cake ------------------------------------------------------- On Hand: cups flour 24 eggs 28 cups sugar 10 How many cakes can be made from these amounts?

What is the “limiting reactant” (LR) What’s left over and how much of it is left over?

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Limiting Reagent

(a)Assume a reactant (any one) is LR.

Calculate stoichiometric amount of product (any product) formed.

(b) Pick another reactant and make it the LR.

Calculate the stoichiometric amount of same product formed.

(c) Whichever reactant gives the smaller amount of product is the Limiting Reagent 30

Limiting Reagent The LR is always used up in a chemical reaction. Everything else is in excess (INXS)

.

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Limiting Reagent

Problem: 4 NH 3 + 5 O 2 4 NO + 6 H 2 O 2.25 g NH 3 mixed with 3.75 g O 2 and allowed to react (a) which is LR?

(b) how many grams NO formed?

(c) how much of excess reactant remains?

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Limiting Reactants

Theoretical Yields The amount of product predicted from stoichiometry taking into account limiting reagents is called the theoretical yield.

The percent yield relates the actual yield (amount of material recovered in the laboratory) to the theoretical yield:

% Yield



Actual yield Theoretica l yield



100

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Theoretical Yield

Problem: C 6 H 6 + Br 2 C 6 H 5 Br + HBr (a) theoretical yield of C 6 H 5 Br when 30.0 g of C 6 H 6 reacts with 65.0 g of Br 2 ?

(b) if actual yield of C 6 H 5 Br is 56.7 g, calculate %yield 34