# CHEM I Chapter 9 Chemical Quantities notes (molar conversions)

```CHEM I Chapter 9 Chemical Quantities
Review Chapter 8
Converting Grams to moles
A sample of hydrogen has a mass of .500 grams how many moles does
this sample contain?
0.500 grams H x 1 mol H
= 0.496 mol H in sample
1.008 g H
Converting moles to grams
9.1
Information given by chemical equations
-Chemical changes: rearrangements of atom groupings that are described by
chemical equations
Consider the balanced chemical equation
CO + 2H2  CH3OH
-This is the smallest possible combination of atoms that still represents
the reaction
-But the equation is still balanced even if we multiply this by 12
12CO + 24H2  12 CH3OH
-We could even use other units to balance the equation
1dozen CO + 2Dozen H2  1 dozen CH3OH
This still would factor down to
CO + 2H2  CH3OH
-We could also use a really large number like 6.02 x 1023
6.02 x 1023CO + 2(6.02 x 1023)H2  6.02 x 1023CH3OH
-But what do we call this number 6.02 x 1023 : mole
So this equation would also be written as
1 mol CO + 2 mol H2  1 mol CH3OH
All the balanced chemical equations that we have done can also be expressed in this
way
C3H8 + O2  CO2 + H2O
Balance this!
C3H8 + 5O2  3CO2 + 4H2O
Interpret in terms of molecules
1 molecule of C3H8 reacts with 5 molecules of O2 to give 3 molecules of
CO2 plus 4 molecules of H2O
Interpret in terms of moles
9.2
1 mol of C3H8 reacts with 5 mol of O2 to give 3 mol of CO2 plus 4 mol of
H 2O
Mole-Mole Relationships
-We can use the chemical equation to predict how much product will come from
a given amount of product.
For example: take this balanced chem.. equation for the decomposition of water
2H2O  2H2 + O2
This would read as 2 mol of water yields 2 mol of hydrogen gas and 1 mol of
oxygen gas
What if we start with 4 mole of water, How many moles of each product would
we get?
Think of it as a ratio
2 H2
=
x H2
2H2O
4H2O
x = 4 mol H2
1O2
2H2O
=
x O2
4H2O
x = 2 mol O2
What have we done to the equation?
2[2H2O  2H2 + O2]
-we multiplied the entire equation by two
4H2O  4H2 + 2O2
What if we were to start with 5.8 mol of water, How many moles of products would this yield?
2 ways to do this
1 the ratio way we did before
2 H2
=
x H2
x = 5.8 mol H2
2H2O
5.8H2O
1O2
=
x O2
2H2O
5.8H2O
x = 2.9 mol O2
Or to make the mole ratios: conversion factors based on the balanced chemical
equation.
Before you make the ratio
2H2O  2H2 + O2
get H2O to 1 by dividing by 2
H2O  H2 + ½ O2
Now multiply by 5.8
5.8H2O  5.8H2 + 2.9 O2
-this second way is the way your book does this problems
or use it as a conversion factor
ex. 5.8 mol H2O x 1 mol O2 = 2.9 mol O2
2 mol H2O
-this is how I like the problem done
Ex. 9.3 Mole Ratio in Calculations page263
C3H8 + 5O2  3 CO2 + 4H2O
Calculate the number of moles of oxygen required to react with exactly 4.30
mol of propane. C3H8 in the reaction
4.30 mol C3H8 x
5 mol O2
= 21.5 mol O2
1 mol C3H8
How many CO2 would be produced, how many H2O?
4.30 mol C3H8 x 3 mol CO2
= 12.9 mol CO2
1mol C3H8
4.30 mol C3H8 x
= 17.2 mol H2O
4 mol H2O
1 mol C3H8
9.3
Mass Calculations
-review for converting moles to masses
given the equations
2Al + 3I2  2 AlI3
We have 35 g of Aluminum What mass of I2 should we weigh out to react
exactly with this amount of aluminum?
2 mol of Al requires 3 mole of I2 our mole ratio is
3 mol I2
2 mol Al
Since the ratio is in moles our amount of Al must also be in moles
Grams of Al x
1 mol Al
= moles of Al
Molar mass of AL
35.0 g Al x 1 mol Al x 3 mol I2
= 1.95 mol I2
26.98 g Al
2 mol Al
Now we know the number of moles I2 required how many grams is that?
1.95 mol I2 x
253.8 g I2 = 495 g I2
1 mol I2
Ex. 9.4 page 265
Given C3H8 + 5O2  3CO2 + 4H2O (make sure its balanced)
What mass of oxygen will be required to react with exactly 96.1 g of Propane?
1st = change to moles
2nd = use mole ratio to convert to other chemical
3rd = use molar mass to convert from moles to grams
96.1 g C3H8
x
1 mol C3H8 x
44.09 g C3H8
5 mol O2
1 mol C3H8
= 10.9 mol O2
10.9 mol O2 x 32.0 g O2 = 349 g O2
1 mol O2
Steps for calculating the masses of reactants and products of chemical reactions
1. Balance the equation
2. convert the masses to moles
3. use mole ratio from balanced equation to convert from moles of one chemical to
another
4. Convert from moles to mass
This process of using the chemical equation to calculate the relative masses of reactants and
products involved in a reaction is called stoichiometry
-the balanced equation describes the Stoichiometry of the reaction
Using Stoichiometry Ex. 9.5 page 268
When solid lithium hydroxide and carbon dioxide gas react the products are solid lithium carbonate
and liquid water. What mass of carbon dioxide gas can 1000 grams of lithium hydroxide absorb?
1st write the equation:
LiOH (s) + CO2 (g)  Li2CO3 (s) + H2O(l)
nd
2 balance the equation:
2LiOH (s) + CO2 (g)  Li2CO3 (s) + H2O(l)
3rd start the problems
1st convert to moles:
1000 g LiOH x 1 mol LiOH/23.95 g LiOH = 41.8 mol LiOH
nd
2 use mole ratio to convert between chemicals
41.8 mol LiOH x 1 mol CO2/ 2 mol LiOH = 20.9 mol CO2
rd
3 convert to grams
20.9 mol CO2 x 44.01 g CO2/ 1 mol CO2 = 920 g CO2
9.4
Calculations involving Limiting Reactant
-In a chemical rxn: the reaction can only proceed as long as enough of all the parts
are present
-for example: if you were making a bologna sandwich
you have 4 pieces of bread
2 pieces of cheese
1 slice of bologna
-How many complete sandwiches can you make? 1
We only have 1 slice of bologna
Ex. CH4 + H2O  3H2 + CO
What mass of water is required to react exactly with 249 grams of methane?
249 g x 1mol/16.04 g = 15.5 mol CH4
15.5 mol x 1 mol H2O/1 mol CH4 = 15.5 mol H2O
15.5 mol x 18.02 g H2O/ 1 mol = 279 g H2O
-This means that we need exactly 279 grams of water before all the methane
is used up.
If we only had 200 grams of water for 249 grams of methane?
-not all of the methane would be consumed
-Water is limiting the amount of products that can be made
-Limiting reactant/reagent: a reactant that is consumed first in
the reaction that limits the amount of products that can
If we had 300 grams of water for 249 grams of methane?
-We would have excess water
-Water would be in excess.
-methane would be the limiting reactant
-Always check to see which reactant is limiting when you are given 2 or
more initial amounts
-Using the mole ratio determine which reactant is limiting
-then use the limiting reactant to compute the amt of product
that is produced
Ex. 9.8 page 277
Nitrogen gas can be prepared by passing ammonia gas over solid copper II oxide at high
temps. The other products of the reaction are solid copper and water vapor. How many grams of N2
are formed when 18.1 grams of NH3 are reacted with 90.4 g CuO?
2NH3 + 3CuO  N2 + 3Cu + 3H2O
18.1 g NH3 x 1 mol / 17.03 g = 1.06 mol NH3
90.4 g CuO x 1 mol / 79.55 g CuO = 1.14 mol CuO
Determine limiting reactant: use the mole ration between the 2 reactants
1.06 mol NH3 x 3 mol CuO/2 mol NH3 = 1.59 mol CuO
We have less mol CuO than we need if 1.59 mol CuO is required to react with 1.06
mol NH3 then the CuO will run out before the NH3 and the CuO is the limiting reactant.
1.14 mol CuO x 1 mol N2/ 3 mol CuO = 0.380 mol N2
0.380 mol x 28.02 g/1 mol = 10.6 g N2
9.5
Percent Yield:
-Theoretical yield: the amount of product that will be made calculated from Stoichiometry
-the amt of reactants predicts the amount of product made
-IT shows the maximum amount of product that can be made from the reactants
-Actual Yield: the amount of product actually obtained from a reaction
- usually done experimentally
-Percent yield: comparison of the actual yield to the theoretical yield
- Actual yield/ theoretical yield x 100% = percent yield
Ex. If a reaction produces 6.63 g of N2 but the predicted amount is 10.6 g of N2 what is the percent
yield?
6.63 g / 10.6 g x 100% = 62.5%
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