Conversion FactorsConversion Factors

Molar mass Molar mass

atomic mass in g = 1 moleatomic mass in g = 1 mole Volume of gas at STPVolume of gas at STP

1 mole gas = 22.4L1 mole gas = 22.4L Mole-mole ratio Mole-mole ratio

coefficients from balanced equationcoefficients from balanced equation Avogadro’s numberAvogadro’s number

6.02 x 106.02 x 102323 molecules = 1 mole molecules = 1 mole

StoichiometryStoichiometry

gg

gg

mole X mole X mole Ymole Y

LLMolecules (gas at Molecules (gas at

STP)STP)

Molar massMolar mass

22.4L=1 mol6.02 x 1023 molecules = 1 mole

For great ion review game www2.stetson.edu/mahjongchem/

2AgNO2AgNO33 + H + H22S AgS Ag22S + S + 2HNO2HNO33

How many moles of AgHow many moles of Ag22S will be produced S will be produced from 3.5 moles of AgNOfrom 3.5 moles of AgNO33

3.5 mole AgNO3.5 mole AgNO33 x x 1 mole Ag1 mole Ag22SS == 1.75 mole Ag 1.75 mole Ag22SS

2 mole AgNO2 mole AgNO33

How many grams of HNOHow many grams of HNO3 3 will form when 2.5g of will form when 2.5g of HH22S react?S react?

2.5g H2.5g H22S x S x 1 mol H1 mol H22SS x x 2 mol HNO2 mol HNO33 x x 63.0g HNO63.0g HNO33

34.1g H34.1g H22S 1 mol HS 1 mol H22S 1 mol HNOS 1 mol HNO33

= 9.2g HNO3

MolarityMolarity

moles of solutemoles of solute

L of solutionL of solution

Dilutions Dilutions VV11MM11 = = VV22MM22

How can we make 500 ml of 1.00 M How can we make 500 ml of 1.00 M HCl from 6.00 M HCl?HCl from 6.00 M HCl?

VV11= ? M= ? M11= 6.00 M = 6.00 M VV22= .500L M= .500L M22= 1.00 M= 1.00 M

VV11(6.00M) = (.500L)(1.00M)(6.00M) = (.500L)(1.00M)

VV11 = = .500L∙M.500L∙M == .0833L or 83.3ml .0833L or 83.3ml 6.00M 6.00M

What is the molarity of 250ml What is the molarity of 250ml solution containing 9.46g solution containing 9.46g

CsBr?CsBr?ggmolemole

9.46g x 9.46g x 1 mole CsBr1 mole CsBr == 0.0444 mole 0.0444 mole CsBrCsBr

213 g CsBr213 g CsBr

250 ml = .250 L250 ml = .250 L

M = M = 0.0444 mole0.0444 mole == 0.178 M CsBr 0.178 M CsBr

.250L.250L

The amount of heat gained or The amount of heat gained or lost depends on the amount of lost depends on the amount of

reactants used.reactants used.2Na2Na22OO22 + 2H + 2H22O O 4NaOH + O 4NaOH + O22 + 215.76 kJ + 215.76 kJHow much heat is released by the How much heat is released by the

reaction of 5.0 moles of Nareaction of 5.0 moles of Na22OO22??

5.0 moles Na5.0 moles Na22OO22 x x 215.76 kJ215.76 kJ == 539.4 kJ 539.4 kJ

2 mole Na2 mole Na22OO22

Heat of reaction (QHeat of reaction (Qrr) is negative for ) is negative for exothermic reactions, but positive for exothermic reactions, but positive for endothermic reactions. So Qendothermic reactions. So Qrr = -539.4 kJ = -539.4 kJ

The decomposition of potassium The decomposition of potassium chlorate is an endothermic reaction.chlorate is an endothermic reaction.

2KClO2KClO33(s) + 89.5kJ (s) + 89.5kJ 2KCl(s) + 3O 2KCl(s) + 3O22(g) (g)

How much heat is absorbed to How much heat is absorbed to decompose 1.50 grams of solid KClOdecompose 1.50 grams of solid KClO33??

1.50g KClO1.50g KClO33 x x 1 mole KClO1 mole KClO3 3 x x 89.5kJ89.5kJ

122.5 g KClO122.5 g KClO3 3 2 mole 2 mole KClOKClO33

= .548kJ= .548kJ

Limiting ReactantLimiting Reactant

1 frame and 3 wheels are needed to create one tricycle

If I have 10 frames and 12 wheels, how many tricycles can I make?

NN22HH44 + 2H + 2H22OO22 NN22 + 4H + 4H22OO

Which is the limiting reactant when Which is the limiting reactant when 0.750 moles of N0.750 moles of N22HH44 is mixed with is mixed with 0.500 mole of H0.500 mole of H22OO22??

If all the NIf all the N22HH44 is used, how much H is used, how much H22OO22 would be needed? would be needed?

0.750 mole N0.750 mole N22HH44 x x 2mole H 2mole H22OO22 = = 1.5 mole H 1.5 mole H22OO22

1 mole N1 mole N22HH44

Do we have that much? No! So HDo we have that much? No! So H22OO22 is limiting is limiting

NN22HH44 + 2H + 2H22OO22 NN22 + 4H + 4H22OO

How much of the excess reactant (in moles) How much of the excess reactant (in moles) remains unchanged?remains unchanged?

0.500 moles H0.500 moles H22OO22 x x 1 mole N1 mole N22HH4 4 = = .250 mole.250 mole

2 mole H2 mole H22OO22

0.750 mole - .250 mole = .500 mole N0.750 mole - .250 mole = .500 mole N22HH44

How much HHow much H22O would be formed?O would be formed?

0.500 mole H0.500 mole H22OO2 x 2 x 4 mole H4 mole H22OO = = 1.00 mole H 1.00 mole H22OO

2 mole H2 mole H22OO22

% yield = % yield = actualactual xx 100% 100% theoreticaltheoretical

If 5.50g of hydrogen reacts with nitrogen If 5.50g of hydrogen reacts with nitrogen to form 20.4g of ammonia, what is the to form 20.4g of ammonia, what is the percent yield?percent yield?

NN22(g) + 3H(g) + 3H22(g) (g) 2NH 2NH33(g)(g)Actual = 20.4gActual = 20.4gTheoretical = Theoretical =

5.50g H5.50g H22 x x 1 mol H1 mol H22 x 2 x 2mol NHmol NH33 x x 17.0g 17.0g NHNH33

2.0g H2.0g H2 2 3mol H3mol H22 1mol NH 1mol NH33

% yield = % yield = actualactual xx 100% 100%

theoreticaltheoretical

% yield = % yield = 20.4g20.4g x 100% = 65.4% x 100% = 65.4%

31.2g31.2g

What is the molecular formula of the What is the molecular formula of the molecule that has an empirical molecule that has an empirical formula of CHformula of CH22O and a molar mass of O and a molar mass of 120.12 g/mol?120.12 g/mol?

Empirical formula mass: 12.0 + 2.0 + Empirical formula mass: 12.0 + 2.0 + 16.0 = 30.0 amu16.0 = 30.0 amu

X = X = 120.12 amu 120.12 amu = 4 = 4 30.0 amu30.0 amu

So CSo C44HH88OO44 is the molecular formula. is the molecular formula.

Determining empirical Determining empirical formulaformula

Find the empirical formula of a compound Find the empirical formula of a compound that contains 53.7% iron and 46.3% sulfur. that contains 53.7% iron and 46.3% sulfur.

% composition % composition mass of 100 g sample mass of 100 g sample moles moles mole ratio mole ratio

53.7g x 1 mol/55.8g = .962 mol Fe53.7g x 1 mol/55.8g = .962 mol Fe

46.3g x 1 mol/32.1g = 1.44 mol S46.3g x 1 mol/32.1g = 1.44 mol S

.962/.962 = 1, 1.44/.962 = 1.5.962/.962 = 1, 1.44/.962 = 1.5

1:1.5 ratio x 2 = 2:3 ratio1:1.5 ratio x 2 = 2:3 ratio

FeFe22SS33