What is stoichiometry?The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products.

1. Atomic Mass2. Avogadro’s Number and the Molar Mass of an

Element3. Molecular Mass4. Percent Composition of Compounds5. Writing and Balancing Chemical Equations6. Amount of Reactants and Products7. Limiting Reagents and Reaction Yield

Atomic MassAtomic mass is the mass of the atom in atomic mass unit(amu).

atomic mass unit - a mass exactly equal to one-twelfth the mass of one carbon-12 atom.Carbon-12 is the carbon isotope that has six protons and six neutrons. Setting the atomic mass of carbon-12 at amu provides the standard for measuring the atomic mass of the other elements.

Average Atomic MassCopper, a metal used in electrical cables and pennies among other things. The atomic masses of its two stable isotopes, Cu(69.09%) and Cu(30.91%), are 62.93 amu and 64.9278 amu, respectively. Calculate the average atomic mass of copper.

FIRST STEP: Convert percents to fractions:69.09%= 69.09/100= 0.690930.91%= 30.91/100= 0.3091

SECOND STEP: Add the contributions together.

average atomic mass = (0.6909)(62.93 amu)+(0.3091)(64.9278 amu)

= 63.55 amu

Avogadro’s Numbermole(mol)- is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon-12 isotope

Avogadro’s number(NA) = 6.0221367 x 10²³

molar mass – the mass (in grams or kilograms) of 1 mole of units 1 mol carbon-12 atoms= 6.022x10²³carbon-12 atoms

Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e di CerretoBorn: August 9, 1776(1776-08-09)Died: July 9, 1856 (aged 79)Institutions: University of TurinKnown for: Avogadro’s Law; Avogadro’s Constant

Avogadro’s Number

Remember:6.02 x 1023 (Avogadro’s Constant)

1 mole of 12C has a mass of 12 g

10

Converting grams to moles.

Determine how many moles there are in 5.17 grams of Fe(C5H5)2.

Goal

= moles Fe(C5H5)2

Given

5.17 g Fe(C5H5)2

Use the molar mass to convert grams to moles.

Fe(C5H5)2

2 x 5 x 1.001 = 10.012 x 5 x 12.011 = 120.11

1 x 55.85 = 55.85

mol

g 185.97

g 185.97

mol0.0278

units match

Grams to atoms

Sulfur is a nonmetallic element that is present in coal. When coal is burned, sulfur is converted to sulfur dioxide and eventually to sulfuric acid that give rise to the acid rain phenomenon. How many atoms are in 16.3 g of S?grams of S mole of S number of S atoms

16.3 g S × 1 mol S × 6.022x10²³ S atoms = 3.06 x 10 ²³ S atoms

32.07 g S 1 mol S

There are 3.06 x 10²³ atoms of S in 16.3 g of S.

Molecular Mass

molecular mass – sum of the atomic masses(in amu) in the molecule

We need to multiply the atomic mass of each element by the number of atoms of that element present in the molecule and sum over all the elements.

Example 3.1 Calculate the molecular mass of the following

compounds: (a) sulfur dioxide(SO₂) and (b) caffeine(C₈H₁₀N₄O₂).

(a)There are two O atoms and one S atom in SO₂, somolecular mass of SO₂ = 32.07 amu + 2(16.00 amu)

= 64.07 amu

b)

molecular mass of C₈H₁₀N₄O₂ = 8(12.01amu) + 10(1.008amu) + 4(14.01amu) + 2(16.00amu)= 194.20 amu

Example 3.2Methane (CH₄) is the principal component of natural gas. How many moles of CH₄ are present in 6.07 g of CH₄?

1 mol CH₄ = 16.04 g CH₄

6.07 g CH₄ × 1 mol CH₄ ₌ 0.378 mol CH₄ 16.04 g CH₄

Percent Composition

percent composition – is the percent by mass of each element in a compound

percent composition= n × molar mass of an element × 100%

molar mass of a compound

where n is the number of moles of the element in 1 mole of the compound

Example 4.1Phosphoric acid (H₃PO₄) is a colorless, syrupy liquid used in detergents, fertilizers, toothpastes, and in carbonated beverages for a “tangy” flavor. Calculate the percent composition by mass of H, P, and O in this compound.

molar mass of H₃PO₄ = 97.99 g

%H = 3 (1.008g) H × 100% = 3.086%97.99 g H₃PO₄

%P = 30.97 g P × 100% ₌ 31.61 % 97.99 g H₃PO₄

%O = 4 ( 16.00 g ) O × 100% ₌ 65.31%97.99 g H₃PO₄

Example 4.2Ascorbic acid (vitamin C) cures scurvy. It is composed of 40.92 percent carbon (C) , 4.58 percent hydrogen(H), and 54.50 percent oxygen (O) by mass. Determine its empirical formula.

Mass percent moles of each element

EMPIRICAL FORMULA mole ratios of elements

nc ₌ 40.92 g C × 1 mol C ₌ 3.407 mol C

12.01 g CnH ₌ 4.58 g H × 1 mol H ₌ 4.54 mol H

1.008 g HnO = 54.50 g O × 1 mol O = 3.406 mol O

16.00 g O

We arrive at the formula:C₃.₄₀₇H₄.₅₄O₃.₄₀₆

C: 3.407 ≈ 1 H: 4.54 ₌ 1.33 O: 3.406 ₌ 1 3.406 3.406 3.406

TRIAL AND ERROR PROCEDURE:

1.33 × 1 = 1.331.33 × 2 = 2.661.33 × 3 = 3.99 ≈ 4

Example 4.3Chalcopyrite (CuFeS₂) is a principal mineral of copper. Calculate the number of kilograms of Cu in 3.71 × 10³ kg of chalcopyrite.

molar mass of Cu = 63.55 gmolar mass of CuFeS₂ = 183.5 g

%Cu ₌ molar mass of Cu × 100%molar mass of CuFeS₂

₌ 63.55 g × 100%183.5 g

₌ 34.63 %

Mass of Cu in CuFeS₂ = 0.3463 × (3.71 × 10³ kg) = 1.28 × 10³ kg

Chemical Reactions and Chemical Equations

chemical reaction – a process in which a substance is changed into one or more new substances

chemical equation – uses chemical symbols to show what happens during a chemical reaction

Writing Chemical EquationsConsider what happens when hydrogen gas (H₂) burns in air(which contains oxygen, O₂) to form water (H₂O). This reaction can be represented by the chemical equation:

H₂ + O₂ H₂O

where the “plus” sign means “reacts with” and the arrow means “to yield”. This symbolic expression can be read as : “Molecular hydrogen reacts with molecular oxygen to yield water”

reactants – the starting materials in a chemical reaction

product – the substance formed as a result of a chemical reaction

Balancing Chemical EquationsSTEPS:1. Identify all reactants and products and write their correct

formulas on the left side and right side of the equation.H₂ + O₂ H₂O

2.Begin balancing the equation by trying different coefficients to make the number of atoms of each element the same on both sides of equation. NOTE: Do not change the subscripts because it would change the identity of the substance.

2H₂ + O₂ 2H₂O

3. Check your balanced equation to be sure that you have the same total number of each type of atoms on both sides of the equation arrow.

Amounts of Reactants and Products

Mass of reactant Moles of reactant

Mass of product Mole of product

Example 6.1All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction as that between lithium and water.

2Li(s) + 2H₂O(l) 2LiOH(aq) + H₂(g)

(a)How many moles of H₂ will be formed by the complete reaction of 6.23 moles of Li with water?

(b)How many grams of H₂ will be formed by the complete reaction of 80.57 g of Li with water?

a) moles of H₂ produced=62.23 mol Li × 1 mol H₂ = 3.12 mol H₂

2 mol Lib) grams of Li moles of Li moles of H₂ grams of H₂

moles of Li = 80.57 g Li × 1 mol Li ₌ 11.61 mol Li 6.941 g Li

moles of H₂ = 11.61 mol Li × 1 mol H₂ ₌ 5.805 mol H₂ 2 mol Li

grams of H₂ = 5.805 mol H₂ × 2.016 g H₂ ₌ 11.70 g H₂ 1 mol H₂

Limiting Reagent and Reaction Yield

stoichiometric amounts – in the proportions indicated by the balanced equation

limiting reagent – the reactant used up first in a reaction

excess reagent – the reactants present in quantities greater than necessary to react with the quantity of the limiting reagent

Limiting Reagent

The limiting reagent is the reactant present in the smallest stoichiometric amount

Limiting Reactants

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

Example 7.1Urea [(NH₂)₂CO] is prepared by reacting ammonia with carbon dioxide:

2NH₃(g) + CO₂(g) (NH₂)₂CO(aq) + H₂O(l)

In one process, 637.2 g of NH3 are treated with 1142 g of CO2.

a)Which of the two reactants is the limiting reagent?b) Calculate the mass of (NH₂)₂CO formed.

c) How much excess reagent (in grams) is left at the end of the reaction?

FIRST STEP: 637.2 g NH3

grams of NH3 moles of NH3 moles of (NH₂)₂CO

moles of (NH₂)₂CO = 637.2 g NH3 × 1 mol NH3 × 1 mol (NH₂)₂CO 17.03 g NH3 2 mol NH3

= 18.71 mol (NH₂)₂COSECOND STEP: 1142 g CO2

grams of CO2 moles of CO2 moles of (NH₂)₂CO

moles of (NH₂)₂CO = 1142 g CO2 × 1 mol CO₂ × 1 mol (NH₂)₂CO 44.01 g CO2 1 mol CO2

= 25.95 mol (NH₂)₂CO

Theoretical YieldThe theoretical yield is the 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, the amount one actually produces and measures

Percent YieldA comparison of the amount actually obtained to the amount it was possible to make

Actual YieldTheoretical YieldPercent Yield = x 100

1. A sample contains 27.1 g of calcium oxide.  How many moles of calcium oxide are in the sample? NOTE: Use the Periodic Table to find the molecular mass (grams per mole)

2. How many atoms are in 0.652 mol of iron? NOTE: A mole is by definition 6.0220 x 1023 particles which can generally be rounded to 6.02 x 1023.

3. How many liters does 3.8 moles of O2 occupy at STP (standard temperature and pressure)? NOTE: At STP, 1 mole of any gas = 22.4L.  STP is 273K (0C) and 1 atm.

4. A solution of NaCl has a molarity of 0.549 M.  How many moles are in 350. mL of this solution? NOTE: Molarity is MOLES per LITER.  The volume in milliliters must be converted to Liters.

5. How many grams of carbon dioxide are there in a container with a volume of 4.50L at STP?

6. How many moles of nitrogen are there in a 7.45 mol sample of (NH4)3PO4?

7.  If 120. g of propane, C3H8, is burned in excess oxygen, how many grams of water are formed?

8. 90.0 g of FeCl3 reacts with 52.0 g of H2S.  What is the limiting reactant?  What is the mass of HCl produced?  What mass of excess reactant remains after the reaction? NOTE: The limiting reactant is the reactant that limits the amount of product that can be formed and is completely consumed during the reaction.  The excess reactant is the reactant that is left over once the reaction has stopped due to the limiting reactant.  See notes limiting reactant.

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GROUP 1 (IV-DIAMOND)Agulan, Arianne Louis

Ca-at, Julienne Rose

Catubig, Fredelyn Eve

De Guzman, Abegail Claire

Africa, Raymart