stoichiometrystudyguide

Stoichiometry Study Guide A.

MOLECULAR WEIGHT (MW) sum of the atomic weights of all the atoms in a molecule of a substance FORMULA WEIGHT (FW) sum of the atomic weights of all the atoms in a formula unit of a substance Note: We don’t always distinguish between MW and FW; i.e., we don’t always

distinguish between molecular compounds and ionic compounds when calculating the molar mass of a substance because the calculation is the same for both types of  compounds  Examples: C2H5OH C = (2) x 12.0 amu H = (6) x 1.01 amu O = (1) x 16.0 amu 46.1 amu (the molar mass is 46.1 grams)

B. 1. a) b) c) 2. a) b)

NaCl Na = (1) x 23.0 amu Cl = (1) x 35.5 amu 58.5 amu (the molar mass is 58.5 grams)

MOLE LE-- oft often en call calleed Av Avogad ogadro ro's 's Numbe umberr (N (NA) Definitions are equivalent: the weight of 6.02 X 10 23 particles expressed in grams

the quantity that contains as many particles as the number of atoms in 12g of  12C the formula weight of a substance expressed in grams; also called molar mass ( MM) Concept of the mole: To weigh out one mole of a substance, weigh out a mass equal to its FW The coefficients in a balanced chemical equation are the number of moles of each substance required for that particular reaction:

CH4 + 2O2 → CO2 + 2H2O This balanced equation means “one mole of methane + two moles of oxygen gives one mole of carbon dioxide + two moles of water”

3.

Think of a balanced chemical equation on microscopic or macroscopic scale: CH4 + 2O2 CO2 + 2H2O → a) [1 molecule] [2 molecule] [1 molecule] [2 molecules] [1 mole] [2 mole] [1mole] [2 moles] b)

4.

Useful conversions that involve the mole are: a)

grams of a substance converted to moles

example:

convert 53 grams of methane to moles of methane:

FW of methane (CH4) = 16.0 amu; MM = 16g/mol moles = sample wt./MM = 53g/16g/mol = 3.3 moles [note: note: grams cancel ]

b)

moles of a substance converted to grams

example:

convert 3.5 moles of sodium dichromate to grams:

FW of Na2Cr 2O7 = 262g/mol grams = (moles of sample)(MM of sample) = (3.5 mol) (262 g/mol) = 917 g [mol cancel] c)

quantity of substance to number of particles in that substance:

recall that one mole is 6.02 X 10 23 particles example:

E = (X 10)  10) 

calc the number of atoms in 7.46g of Li convert grams to moles to number of atoms:

atom atomss of Li = (7.4 (7.46g 6g Li)

(1 mol mol Li) (6.9 (6.914 14 g Li)

The student should be able to do the following conversions: grams to moles g → mol • grams to #atoms g → mol → #atoms • •

moles to grams moles to #atoms

mol → g mol → #atoms

•

#atoms to moles

#atoms → mol

•

#atoms to grams

#atoms → mol

•

(6.0 (6.02E 2E23 23 atom atomss of Li) = 6.47E23 (1 mol of Li)

(g of sample) / (MM) (g of sample/gmw) X (NA) (mol of sample) X (MM) (mol of sample) X (NA) (#atoms of sample/NA)

→

g

(#atoms of sample/NA) X (MM)

C.

STOICH STO ICHIOM IOMETR ETRYY- quanti quantita tativ tivee relat relation ionshi ships ps bet betwee ween n react reactant antss and produc products ts

1.

Stoichiometry of one substance to another in a balanced chemical equation

a)

determine how much of  reactant gives a particular amount of product

example: "how many grams of calcium phosphate, Ca 3(PO4)2, are needed to give 5g of phosphorus (P 4)?" here is the balanced equation: 2 Ca3(PO4)2 + 6 SiO2 + 10C → 1P4 + 6 CaSiO3 + 10 CO From the coefficients it is determined that: 2 moles of Ca3(PO4)2 will give 1 mole of P4 The question asks for grams, so moles must be converted to grams at some point in the calc. The "roadmap" of the calc might look something like this: 5g of P 4 The calculation is: 5g P4

1 mol P4 123.88g of P 4

→

mol P4

→

mol Ca3(PO4)2

2mol of Ca 3(PO4)2 1mol of P4

→

g of Ca3(PO4)2

310.18g of Ca3(PO4)2 1mol of Ca3(PO4)2

=

25.04g of Ca3(PO4)2

note: all units cancel except for the desired unit, g of phosphate [in the form of Ca3(PO4)2] b)

determine how much product from a particular amount of reactant

The same question could have been asked "backwards"; i.e. "how much P 4 is produced from 25.04g of phosphate?" phosphate?" The "roadmap" is similar (except backwards) and the calc is analogous: analogous: 25.04g of Ca 3(PO4)2

1mol of Ca3(PO4)2 310.18g of Ca 3(PO4)2

1mol of P 4 2mol of Ca3(PO4)2

123.88g of P 4 = 5 g P4 1 mol P 4

A sample of N2O5 produced 1.618g of O 2; how many grams of NO 2 formed? The balanced chemical equation is: 2N2O5 → 4NO2 + O2 From the coefficients: for every 1mole of O2 produced, there are 4 moles of NO2 produced

c)

The "roadmap" might look like this: g of O2 → moles of O2 → moles of NO2 And the calc: 1.618g O 2

1mol O2 32.00g O 2

4mol NO2 1mol O2

→

g of NO2

45.99g NO 2 1mol NO2

= 9.301 g NO2

2. Limiting Reagent- the reagent that is completely consumed in a reaction; the other reactants which are not completely "used up" are said to be "in excess"

A typical problem using the concept of limiting reagent asks for the quantity of reagent A that is required to consume reagent B (or vice versa). Starting with a balanced chemical equation such as this one: CS2 + 3O2

→

CO2 + 2SO2

consider the following:

3 moles of O 2 are required to completely "use up" 1 mole of CS 2 and/or  1 mole of CS 2 is needed to completely consume 3 moles of O 2 Problem:

15.0g of CS 2 and 35g of O 2 are allowed to react; which is the limiting reagent?

Solution:

1st- co convert ma masses to to mo moles 2nd- establish the molar relationship between each reactant and one of the  products by using the coefficients from the balanced equation (here we use SO SO2 product but CO 2  product would also work) 15.0g/7 15.0g/76g/ 6g/mo moll = (0.197m (0.197mol ol CS2) (2mol SO2)

= 0.394mol SO2

(1mol CS2) 35g/32 35g/32.0g .0g/m /mol ol = (1.094 (1.094 mol mol O 2)

(2mol SO2)

= 0.73mol SO2

(3mol O2)

CS2 is the limiting reagent re agent because the amount given, 15.0g, produces only 0.394 moles of  (SO2); whereas, the amount of O2 given, 35g, produces 0.73 moles of product (SO 2).  Note: The moles of product (in this example, 0.394mol SO SO 2) are always determined by the moles of the limiting reagent (0.197mol CS 2)! Alert! Problem-solving that involves a limiting reagent is almost always recognized by recognized by the fact that more than one reagent is given in the problem. For example: A solution containing 15 g of AgCl and a solution containing c ontaining 25 moles of NaBr  were mixed. How much product is theoretically formed?

3.

% Yield - amount of product obtained after reaction "goes to completion"

% yld = (actual yld) X 100% (theor. yld) Actual yld is the amount obtained at the end of the experiment. Theoretical yld is the amount calculated from the stoichiometry of the reaction.