Regents Chemistry Stoichiometry

Regents Chemistry

StoichiometryStoichiometry

Atomic Mass Unit (amu)

• amu- unit used to for mass of an atom• amu of Oxygen? 16amu

• Why not in grams? – Mass of oxygen is really 2.7 × 10-23g– Atoms are too small, number is too bulky

02

Find the mass of the following atoms

1.Mg = ______

2.Li = ______

3.Cl = ______

4. Al = ______

5. Ca = ______

6. H = ______

24 amu

7 amu

35 amu

23 amu

20 amu

1 amu

Types of Elements

Monoatomic Element

•One atom of an element that is stable enough to stand on its own (VERY RARE)

– not bonded to anything

Diatomic Elements

•Elements whose atoms always travel in pairs (Br,I,N,Cl,H,O,F)

– Bonded to another atom of the same element.

Formula Mass

• mass of an atom, molecule, compound

• O2

• This means that mass of O2 = 2 × _______ amu = ________ amu

• Notice how we use rounded numbers

Subscript = tells you the total number of atoms in the compound/molecule.

16 32

Gram Formula MassGram Formula Mass

FormulasFormulas

• Molecular – actual ratios of atoms in a molecule or compound.– ex: C4H10 or C6H12O6

• Empirical – simplest ratio of atoms in a compound– ex: C2H5 or CH2O

• Structural – shows the arrangement of the atoms in the actual compound

ExampleExample

• The empirical formula is CH and the molecular mass is 26. What is the molecular formula?

1.C2H2

2.C3H3

3.C4H4

Step 1: find mass of empirical

CH = 13amu

Step 2: Divide molecular mass by empirical mass

26 (given)/13 = 2

Step 3: multiply empirical formula by answer in step 2

CH × 2 = C2H2

ReviewReview- The molecular formula is C3H6. What is the empirical formula?

______________

- Which is an empirical formula?

1.C2H2

2.H2O

3.H2O2

4.C6H12O6

- What is the empirical formula of the compound whose molecular formula is P4O10?

1.PO

2.PO2

3.P2O

4.P8O20

Chemical EquationChemical Equation• Shows which bonds are broken and which

bonds are built. – Numbers of atoms on the left side must equal

number of atoms on the right side of the arrow– After the elements are correctly written, only

the coefficient can be changed. – No coefficient means there is only one

molecule

H2 + O2 H202 atoms H 2 atoms O 2 hydrogen atoms

1 oxygen atom

24 =

2 =

24 =

Fix this equation!Fix this equation!• Formation of salt from sodium and

chlorine gas.

Na + Cl2 → NaCl

Na Cl ClNa ClPRODUCES

Modeling Conservation of Modeling Conservation of MatterMatter

Review equationsReview equationsMg + Cl2 MgCl2Ca + HCl CaCl2 + H2

Ca + H20 Ca(OH)2 + H2

Given the incomplete equation: 2N2O5(g)

Complete the balanced equation. 1.2N2(g) + 3H2(g)

2.2N2(g) + 2O2(g)

3.4NO2(g) + O2(g)

4.4NO(g) + 5O2(g)

Reaction TypesReaction Types

Four Basic Types

1.Synthesis

2.Decomposition

3.Single replacement (substitution)

4.Double replacement

SynthesisSynthesis

• Formation of only ONE product from two reactants, but not always.

Examples:

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

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) because O2 combines with both metal and nonmetal to form two oxides.

DecompositionDecomposition

• One reactant breaks apart to form several products.

• AKA combustion if products are CO2 and H2O

Examples:

2H2O2(aq) 2H2O(l) + O2(g) hydrogen peroxide decomposes over time to leave behind water.

Single ReplacementSingle Replacement

• A more active metal replaces a less active metal in a compound, or vice versa.

Example

2Fe(s) + 6HCl(aq) 2FeCl3(aq) + 3H2(s)

what happens when a metal becomes corroded by an acid where an element is reacting with a compound.

Double replacementDouble replacement

• Reaction between aqueous compounds.

• Cations and anions switch position.

• If an insoluble precipitate forms, the reaction is an end reaction, otherwise an aqueous mixture of ions

ExampleAgNO3 (aq) + NaCl (aq) NaNO3(aq) + AgCl(s)

ReviewReview

1. Cl2(g) + 2NaBr(aq) 2NaCl(aq) + Br2(l)

2. FeCl3(aq) + 3NaOH(aq) Fe(OH)3(s) + 3NaCl(aq)

3. 2Mg(s) + O2(g) 2MgO(s)

4. H2CO3(aq) H2O(l) + CO3(g)

SRSR

DRDR

SS

D/CombustionD/Combustion

Gram Formula Mass (GFM)Gram Formula Mass (GFM)

AKA

•Molar Mass

•Atomic mass

•Molecular mass (only in covalent)

•Mass of formula in grams

Formula MassFormula Mass

• Sum of atomic masses in the molecule

• What is the formula mass (or molecular mass) of K2CO3?

atom Total of each

mass total

K 2 x 39 = 78

C 1 x 12 = 12

O 3 x 16 = 48

Total = 138

gram formula mass (GFM)gram formula mass (GFM)

• GFM – describes the mass of one mole of a compound– To find GFM, add individual GAM for each

element in the compound• H20 = 1.00 + 1.00 + 16.00 = 18.00 g

• (NH4)2SO4 = N = 2 × 14 = 28

H = 8 × 1 = 8

S = 1 × 32 = 32

O = 4 × 16 = 64

132 g

Gram formula Mass (molar Gram formula Mass (molar mass)= mass in gramsmass)= mass in grams

• Mass of 6.02 x 1023 particles (1 mole of particles).

• If you weigh 6.02 x 1023 particles (1 mole of particles) of K2CO3, the weight on the scale will be 138 grams

Compound Formula mass, amu

Mass of 6.02 x 1023 particles

(on scale)

K2CO3 138 138g

H2O 18 18g

O2 32 32g

Molecular MassMolecular MassH20 = 1.01 + 1.01 + 16.00 = 18.02 g/mol

(NH4)2SO4 = N = 2 × 14.00 = 28

H = 8 × 1.01 = 8.08

S = 1 × 32.00 = 32

O = 4 × 16.00 = 64

132.08 g/mol

Determine the molecular Determine the molecular Formula of the followingFormula of the following

HNO3

H N O

1×1 1×14 3×16

1 + 14 + 48

63amu

(NH4)2CO3

N H C O

2×14 8×1 1×12 3×16

28 + 8 + 12 + 48

96amu

% Composition% Composition(NH4)2CO3 96amu96amu

N H C O

28/96 8/96 12/96 48/96

×100 ×100 ×100 ×10029.17% 8.33% 12.50% 50.00%

Needs to total 100%

HNO3 63amu63amu

H N O

1/63 14/63 48/63

×100 ×100 ×100

1.59% 22.22% 76.19%

Needs to total 100%

% Given instead% Given instead

• Cmpd is 86% C and 14% H. What is the empirical formula?

C H

86/12 14/1

7.17 C 14 H

7.17/7.17 14/7.17

1 2

CH2CH2

% of water in Na2CO3 • 10H2O (formula mass = 286)?

H2O = 2 + 16 = 18amu

10 × 18am = 180

180/286 × 100 = 62.94%62.94%

RuleRule

1 mole (of molecules)1 mole (of molecules)

EqualsEquals

1 gram molecular mass1 gram molecular mass

EqualsEquals

6.02 x 106.02 x 1023 23 molecules molecules

Equals Equals

22.4 liters (for gases at STP)22.4 liters (for gases at STP)

ExampleExample• What is the formula mass of NO2?

46 gram

• What is the mass of 2 moles of NO2? 2 X 46 gram = 92 grams

• What is the mass of 12 x 1023 molecules of NO2? 12 x 1023 / 6.02 x 1023 = 2

2 x 46 = 92 grams

• What is the mass of 44.8 liters of NO2? 44.8 / 22.4 = 2 2 x 46 = 92 grams

DensityDensity

Density = mass / volume

Usually expressed for gases in grams/liter

Gram formula mass = density at STP (g/L) x 22.4 liters

A piece of aluminum has a mass of grams and a volume of ml. What is its density?

1.35 g/l x 22.4 l = 30.24 grams

Another exampleAnother example

Which gas has a density of 1.70g/l at STP?

1.F2

2.He

3.N2

4.SO2

GFM = density (g/l) x 22.L

= 1.7 g/l x 22.4 liters

GFM = 38.08 grams

Total the mass of each choice to find the answer.

Mole-Mole ProblemsMole-Mole Problems• Answers how many moles

of one element or compound react with a given number of moles of another element or compound.

• How many moles of Ca are needed to react completely with 6 moles of H2O in the following reaction:

Step 1: Balance equation

Step 2: Cross out molecules not needed.

Step 3: Write mole number on top of given formula and an x on the unknown

Step 4: Write mole number on bottom of formula from balanced equation

Step 5: set up proportions

Ca + HCa + H22O O Ca(OH) Ca(OH)22 + H + H2226x

1 mole 2 mole

Review/PracticeReview/Practice

• Given the reaction: CH4 + O2 CO2 + H2O– How many moles of oxygen

are needed for the complete combustion of 3.0 moles of CH4?

1.6.0 moles

2.2.0 moles

3.3.0 moles

4.4.0 moles

• What amount of oxygen is needed to completely react with 1 mole of CH4?

1. 2 moles

2. 2 atoms

3. 2 grams

4. 2 molecules

Avogadro’s numberAvogadro’s number

• measured to be approximately 6.022 x 1023 (to 4 s.f)

• Chemists use the mole in the same way that grocers use the dozen for groups of 12 and stationers use the ream for groups of 500. 

•  we can use the mole without being overly concerned about exactly how many objects it represents

MoleMole• smallest measurable mass of matter contains

trillions of atoms, so chemists use a unit of amount called the mole (abbreviated mol).– one mole is the number of atoms in 12 g of carbon-

12• one atom of tin-120 has a mass of 120 u, it follows that one

mole of tin-120 atoms will have ten times the mass of one mole of carbon-12 atoms, i.e. 120 g. 

– the mass in grams of one mole of atoms of any element will be numerically equivalent to its atomic mass in g/mol.

DIMODIMO

Dimensional AnalysisDimensional Analysis

Review Review

• Which gas sample contains a total of 3.0 x 1023 molecules?

1. 71g of Cl22. 2.0 g of H2

3. 14g of N2

4. 38g of F2

A sample of an unknown gas at STP has a density of 0.630 g/l. What is the gram molecular mass of this gas?

1. 2.81g

2. 14.1g

3. 22.4g

4. 63.0g

• Which quantity represents 0.500 mole at STP?

1. 22.4L of Ar

2. 11.2L of N2

3. 32.0L of H2

4. 44.8L of He

Using Avogadro’s numberUsing Avogadro’s number

• how many atoms are in a sample of silicon that has a mass of 5.23 g. 

• Moles can be number of atoms or particles in a molecule

5.23 g silicon 

x 1 mol silicon =0.186 mol

silicon

    28.09 g silicon    

0.186 mol silicon

x 6.022 x 1023 atoms silicon =1.12 x 1023 atoms

silicon

    1 mol silicon                 

MoleMole

– mass of 1 atom = mass of a mole of atoms / 6.022 x 1023

– mass of 1 C atom = 12.01 g / 6.022 x 1023 C atoms

• mass of 1 C atom = 1.994 x 10-23 g

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