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Unit 2: Quantities in Chemistry

Section: 2.1 – 2.14Pages 80-163

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Mass, Moles, & Molar Mass

Term Definition

Isotopic Abundance Relative quantities of isotopes in a natural occurring element (%)

E.g. Carbon has 2 isotopes C-12 and C-13. Of Carbon’s two isotopes, there is 98.9% C-12 and 11.1% C-13. Find the average atomic mass of Carbon.

Term DefinitionFormula mass unit The mass of one molecule of an ionic compound

in a.m.u.

E.g. Calculate the formula mass of CaCl2.

averageamtoangsic percent of isotope 1 mass of isotope1 t

Idivide by 100 to get it as a decimal

AVG C R mass of C R t C B mass of C Datomicmass O 989 I 2 t o 0111 B

12 01 am u

ooo the average atomic mass of C is 12.0 I am

find the masses from the periodictable

Coy 2

I 40.078 t 2 35,453

110 98 t am u

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The Mole

Molar Mass, M The mass, in grams, of one mole of a chemical entity

Term Definition

Mole 6.023 x 1023 entitiesAvogadro’s Constant, NA The number of entities in one mole, 6.023 x

1023

E.g. Calculate the molar mass of NaCl.values are from the PT

M Naa INIT9907 375358 t't3g mole

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Number & Amount of Entities

How many oxygen molecules, O2(g), are in a cylinder that contains 48.0 g of oxygen ?

m

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Calculations Involving the Mole

n

m

M

E.g., Calculate the mass, in grams, of 2.00 moles of calcium atoms.

E.g. What amount of gold is in a 275.8g nugget of pure gold? And how many atoms does this represent?

n

N

NA

mass g ofparticlemotM01 y ymolarmass 6,021023

glmol

Ym nM2,00mm 40.078gal

KE 80.16gooo 2.00moksofcaatomshaveamaesof80bg3

Yn I f N nNAM litomol 6.02 1023275,8g fig 3 102 atoms796.97galElettomolofAu 80275.8g ofpuregold2 VFepresents 8.43 18atoms

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Calculations Involving the Mole Continued.

Calculate the mass of 1 mol of sodium hydrogen carbonate (baking soda), NaHCO3.

Sodium fluoride is added to toothpaste and tap water to prevent tooth decay. Calculate the mass of 2.00 mol of sodium fluoride, NaF(s)

How many water molecules are in a 25.0g sample of water, H2O(l)?

Mwattco5 119907 1 079 tfaz.com t 03C is aaa

if 84.01gImo IM n M ooo l mo I of NaHC03Z

l 84.01 it

4 84,01g has a morass of 84.01g

MnarjIYE.no ticks 998F 4499glmol

KM n M

127141.9983.98g

4082 mot of Nat has a mass of 83.98g1 convert mass to moles2 convert moles to molecules

In Hp IM IN n NA25.0g 1388m01 6.02 1023

18.01g1m01 HE 8.36 1023molecules

III 388m01 HO ooo25.0g of water1 represents 8.36 1023molecule

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Determining Empirical FormulasTerm Definition

Law of Constant Composition

A compound contains elements in certain fixed proportions. E.g. NaCl, H2O, C6H12O6

Percentage Composition

The percentage, by mass, of each element in a compound.

Empirical Formula A formula that gives the lowest ratio of the atoms in a compound

The percentage composition of a compound was found to be 69.9% iron and 30.1% oxygen. What is the empirical formula of the compound.

Step 1: Percent to Mass: Calculate Mass (m) of each element in a 100 g sample ” Assume a 1 mol sample”

Step 2: Mass to Mole: Convert Mass (m) into Amount in moles (n)

Step 3: Divide by small. Divide ALL mole answers in step 2 by smallest value.

Step 4: Multiply ‘til whole. If any answer in step 3 ends with a .5 then multiply all answers in step 3 by “2”

4 Assume a 100g sampleMee 69.9g mo 30.1g

zNfe M_M no Im69,9g 30.1g55.8445 15.999gImoI

3E1 252m01Fe 1 881m01 0

If g1.88151,2522xtl91 grgytiplyzbothTfofhe.E

F is Feroz

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Determining Molecular FormulasTerm Definition

Molecular Formula

A formula that indicates the actual number of atoms in one molecule of a compound.

The empirical formula of a compound is CH3O, and its molar mass is 93.12 g/mol (determined by a mass spectrometer). What is the molecular formula of the compound?

Step 1: List given values.

Step 2: Determine Molar Mass of the Empirical Formula.

Step 3: Determine the multiple! A ratio of the Molar Mass of Compound to Molar Mass of Empirical Formula.

Step 4: Calculate Molecular Formula. Apply the multiple to all subscript numbers in the EF.

MD

EF is CH30 Mme 93.12gImoIen

8133410.979 Eisman

2multiple Molar Mass of the compound FMolar mass of the EF93 12gImoI

i 03g Imo1

kt ooo the MF is CzHqOz

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Determining Percent Composition

Calculate the percentage composition of potassium sulfate, K2SO4.

Step 2: Calculate the Total Mass of Each Element in the Compound.

Step 3: Calculate Molecular Mass (or formula unit mass) of Compound.

Step 3: Calculate Percentage Composition by Mass of Compound.

Step 1. If given a formula only you must “Assume a 1 mol sample”

I Assume a Imd sample of Kiso'tMk 2 39.098 Ms 1132,0657 mo

32,065g445,999

78.20g 64.00g

MKz5OH 2 39,098 t 1 32.0657 4 5.999

174,26g

f k mmE x 00 5 mmsasoyxioo 0emm.jo o

78.2Ogxl00 3ff.O659gx00 6409174.26g 174.26

I 4.80 4 Vs 0 18.40 5 36.800

ooo the compound is 44.80 1418.405

and 36.80 0

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% Concentration V/V, m/VTerm Definition

Quantitative analysis

Measurement of a quantity of a chemical entity.

Concentration, C

A ratio of the quantity of solute in a solution.

A salt solution is formed by mixing 2.80 g of NaCl(s), in enough water to make exactly 250 mL of solution. What is the m/v percentage concentration of sodium chloride salt solution?Step 1: List Given Values.

Step 2: Write Percentage Concentration Equation, Substitute Values,

& Solve.

M Nacl 2,80g solution _250mL

MIU M solute g tooV solution ML

2.80g250mL

X100

k l 123ooo the m V of the solution is 1.12

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Molar Concentrationc

n V

c – is the molar concentration in mol/L.

n – is the amount of solute in moles.

V – is the volume of the solution in L.

A sodium hydroxide solution contains 0.186 mol of sodium hydroxide in 0.250 L of solution. Calculate the molar concentration of the sodium hydroxide solution.Step 1: List Given Values.

Step 2: Write Molar Concentration Equation, Substitute Values, &

Solve.

NNaOH 0 186 mo l V 0.250L C

Vic F0.186motO 250 L

2 0 744MOHL

o the concentration ofthe solution is

3 0.744 molar

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Parts Per MillionTerm Definition

Parts Per Million Concentration unit that is used for very low concentrations; one part solute for every million parts of solvent.

In a chemical analysis 2.2 mg of oxygen was measured in 250 mL of pond water. What is the concentration of oxygen in ppm?

Step 1: List Given Values.

Step 2: Write Percentage Composition Equation.

Step 4: Substitute Values into Equation and Solve.

ppmMsolute 106Msolutiong

DHzo Ig mL

Moz 0.002g MHP 250g so 250mL250g

ppm MsolutecopM solution g

X 106

ppm 6.0022 106250

z 8 ppm

there are 8 ppm in the water

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Concentrations of Solutions Summary!

Mmm

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Diluting Aqueous SolutionsTerm Definition

Dilution The process of decreasing the concentration of a solution by adding more solvent.

C1V1 = C2V2C1 – initial concentration

V1 – initial volume

C2 – final concentration

V2 – final volume

Calculate the final volume of a hydrogen peroxide solution if water is added to a 100 mL of 6% V/V hydrogen peroxide solution until it reaches a volume of 250 mL.Step 1: List Given Values.

Step 2: Write Dilution Equation.

Step 3: Isolate Unknown Value on Left-Hand Side.

Step 4: Substitute Values in & Solve.

VI 100mL C _6 Vz 250mL CE

C V _CzVz

v CECIL isCz 6 1007 so the concentration

2507 of the diluted2.4 solution is2iH

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StoichiometryTerm Definition

Molar Ratio The ratio of amount, in moles, of reactants and products in a chemical reaction.

Stoichiometry Mathematical procedures for calculating the quantities of reactants and products involved in chemical reactions.

Propane, C3H8(g), is a gas that is commonly used in barbecues. Calculate the mass of oxygen that is required to burn 15.0 g of propane.

Step 1: Balance Equation, List Given Values and Molar Masses.

Step 2: Convert Mass of a Given Substance to moles.

Step 3: Convert Amount of a Given Substance to Amount Required

of a Given Substance using a MOLAR RATIO.

Step 4: Convert Amount of Required Substance to Required

Value.

Step 5: Write a therefore statement that answers the question.

Vi GH81g t 5 02cg 302cg t't HzOgm 15.0g me

M t't.ly molM 32i0gmsl

znczHsEFMs.o

3 FIIgHOmol GH 8

I mo l CzHg 5m01 OzO 340m01 X

f X E 1.701 mol Ozm nM

1,701 mo 32.07 E 5 t t g

5titty of 02cg is required

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Summary of Stoichiometry!

g

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Limiting & Excess ReagentsTerm Definition

Limiting Reagent

The reactant that is totally consumed in a chemical reaction.

Excess Reagent The reactant that is present in more than the required amount in a chemical reaction.

Table salt, NaCl(s), can be formed by the reaction of sodium metal with chlorine gas:

2Na(s) + Cl2(g) 2NaCl(s)

A reaction mixture contains 45.98 g of sodium and 142. 0 g of chlorine. Calculate the mass of sodium chloride that is produced.

Step 4: Calculate the Amount of Product.

Step 6: Calculate the Mass of Product.

Step 1: Balance Equation, List Given Values and Molar Masses.

Step 2x2: Convert Mass of a Given Substance to moles.

Step 3x2: Convert Amount of a Given Substance to Amount Required of a Given Substance using a MOLAR RATIO.

You MUST compare to a the same product

2 Na ca t cheap I NaCl S7mm 9aqogimamm.FIa9glmolFiI58ittt3g1mol

zhNa Mff Nch_Mph45.98 142.022.990 70.91

3 2m01 2m01

least 2 moi Na 2m01NaCl

iiieenoltEII.iniEanio E.Eis6 8 pepltAN

9Mnacl nMusing LR

Imd 58 r't't340 58 443g

tooo 58.443g of NaCl is expected

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Percentage YieldTerm Definition

Yield The amount of product produced in a chemical reaction.Actual Yield The amount of product that is actually produced in a

chemical reaction.Theoretical Yield

The amount of product expected from a balanced chemical equation.

Percentage Yield

Actual vs. Theoretical Yield expressed as a percentage of Theoretical Yield.

The most common ore of Arsenic is FeSAs(s), can be heated to produce Arsenic, As(s):

FeSAs(s) FeS(s) + As(s)

When 250 kg of this ore was processed industrially, 95.3 kg of Arsenic was obtained. Calculate the percent yield of Arsenic.

Step 1: Balance Equation, List Values & Molar Masses.

Step 2: Convert Mass of Given Substance to Amount of Substance (n)

Step 3: Convert Amount of Substance to Amount of Required Sub.

Step 4: Convert Amount of Required Sub to Mass of Req. Sub.

Step 5: Calculate Percent Yield.

TRSAS resat AMESM 250000gM 162.83gImd M74.92291mi

InfesAs IM fn 4535 34m01 FesAs

250000g 3162.83glmol

I mol of FesAs 1m01As41535.34mL X

X 1535 34m01 AS

m nM I1535,34 74,922115030g f 115kg

Vgf Y Aff 100 4 9,5 39 woo

t 82.9

To ooo95.3kg represents a 82.9 yield

co J C