semester 1 - weebly
TRANSCRIPT
Stoichiometry Honors chemistry – Semester 1
Order up! 1 bun + 4 tomato slices + 6 onion slices + 2 meat patties + 2 lettuce leaves 1 double hamburger
To make 6 double hamburgers, how many tomato slices do I use? With 24 onion slices, how many double hamburgers can I make? I have 8 lettuce leaves and six meat patties. Will I have ingredients left over? Which one? How much?
What is stoichiometry?
Ratios of reactants and products
• Tells you how much of each:
• Reactant you need
• Product you will make
• Reactant is left over (if any)
Coefficients and Moles
Coefficients in balanced equation show number of molecules, atoms, or ions
Coefficients also show number of moles
2C8H18 + 25O2 → 16CO2 + 18H2O
Mole ratio
Relative number of moles of each substance called the mole ratio
2C8H18 + 25O2 → 16CO2 + 18H2O
Ratio of
• octane to oxygen is 2:25 moles
• carbon dioxide to water is 16:18 (8:9) moles
• octane to carbon dioxide is 2:16 (1:8) moles
Practice
Given:
CH4 + 2O2 → CO2 + 2H2O
What are the mole ratios of:
CH4 to H2O
O2 to CO2
O2 to H2O
1:2
2:1
2:2 (1:1)
Example
To make ammonia, a factory uses formula
N2 + 3H2 → 2NH3
How many mol H2 are used to make 312 mol NH3?
The ratio of H2 : NH3 is 3:2
𝟑𝟏𝟐 𝐦𝐨𝐥 𝐍𝐇𝟑 × 𝟑 𝐦𝐨𝐥 𝐇𝟐
𝟐 𝐦𝐨𝐥 𝐍𝐇𝟑 = 𝟒𝟔𝟖 𝐦𝐨𝐥 𝐇𝟐
Practice
1.34 mol H2O2 decompose: 2H2O2 → 2H2O + O2
• How many mol of H2O form?
• How many mol of O2 form?
Ratio of H2O2 : H2O = 1:1
𝟏. 𝟑𝟒 𝐦𝐨𝐥 𝐇𝟐𝐎𝟐 × 𝟏 𝐦𝐨𝐥 𝐇𝟐𝐎
𝟏 𝐦𝐨𝐥 𝐇𝟐𝐎𝟐
= 𝟏. 𝟑𝟒 𝐦𝐨𝐥 𝐇𝟐𝐎
Ratio of H2O2 : O2 = 2:1
𝟏. 𝟑𝟒 𝐦𝐨𝐥 𝐇𝟐𝐎𝟐 × 𝟏 𝐦𝐨𝐥 𝐎𝟐
𝟐 𝐦𝐨𝐥 𝐇𝟐𝐎𝟐
= 𝟎. 𝟔𝟕𝟎 𝐦𝐨𝐥 𝐎𝟐
Using mole ratio
Four Steps
1) Find mole ratio between chemicals A and B
2) Convert given amount of A into # mol A
3) Use mole ratio to find # mol B
# 𝒎𝒐𝒍 𝑨 ×𝒎𝒐𝒍 𝑩
𝒎𝒐𝒍 𝑨= # 𝒎𝒐𝒍 𝑩
4) Convert # mol B to desired units
Using mole ratio
Use mole ratio to compare amounts of any two compounds in a chemical equation
Substance A Substance B
Grams Grams
Volume Volume
# Moles # Moles
# Particles # Particles
# 𝒎𝒐𝒍𝒆𝒔 𝑨 × 𝒎𝒐𝒍 𝑩
𝒎𝒐𝒍 𝑨= # 𝒎𝒐𝒍𝒆𝒔 𝑩
Step 1. Find mole ratio
Step 2. Convert quantity of A into moles Step 3. Find # moles of B
Step 4. Convert # mol of B into desired units
Example
NASA uses 2.53x104 kg of H2 to launch a rocket using the reaction 2H2 + O2 → 2H2O
• How many kg of O2 does
NASA need for the rocket?
Example
2.53x104 kg of H2 2H2 + O2 → 2H2O
• How many kg O2 does NASA need for the rocket?
1. Compute mole ratio between H2 and O2
2. Convert kg H2 to mol H2
3. Use mole ratio to compute mol O2
4. Convert mol O2 to kg O2
202,400 kg 2.0x105 kg O2
Practice
How many grams N2 are needed to make 1.00x103 g NH3 using the reaction N2 + 3H2 → 2NH3 ?
Mole ratio N2 : NH3 = 1:2
𝟏. 𝟎𝟎𝒙𝟏𝟎𝟑 𝒈 𝑵𝑯𝟑 ×𝟏 𝒎𝒐𝒍 𝑵𝑯𝟑
𝟏𝟕. 𝟎 𝒈 𝑵𝑯𝟑
×𝟏 𝒎𝒐𝒍 𝑵𝟐
𝟐 𝒎𝒐𝒍 𝑵𝑯𝟑
×𝟐𝟖. 𝟎 𝒈 𝑵𝟐
𝟏 𝒎𝒐𝒍 𝑵𝟐
= 𝟖𝟐𝟒 𝒈 𝑵𝟐
Example
Substance A Substance B
Grams Grams
Volume Volume
# Moles # Moles
# Particles # Particles
# 𝒎𝒐𝒍𝒆𝒔 𝑨 × 𝒎𝒐𝒍 𝑩
𝒎𝒐𝒍 𝑨= # 𝒎𝒐𝒍𝒆𝒔 𝑩
Example
Given 56 mL POCl3, find the volume of H3PO4 that forms from the reaction
POCl3 + 3H2O → H3PO4 + 3HCl
dPOCl3 = 1.67 g/mL
dH3PO4 = 1.83 g/mL
Step 1. Mole ratio between POCl3 : H3PO4 = 1:1
Example
Step 2. Find # moles of POCl3
𝟓𝟔 𝒎𝑳 𝑷𝑶𝑪𝒍𝟑 ×𝟏. 𝟔𝟕 𝒈 𝑷𝑶𝑪𝒍𝟑
𝟏 𝒎𝑳 𝑷𝑶𝑪𝒍𝟑
×𝟏 𝒎𝒐𝒍 𝑷𝑶𝑪𝒍𝟑
𝟏𝟓𝟑. 𝟓 𝒈 𝑷𝑶𝑪𝒍𝟑
= 𝟎. 𝟔𝟐 𝒎𝒐𝒍 𝑷𝑶𝑪𝒍𝟑
Step 3. Find # moles of H3PO4
𝟎. 𝟔𝟐 𝒎𝒐𝒍 𝑷𝑶𝑪𝒍𝟑 ×𝟏 𝒎𝒐𝒍 𝑯𝟑𝑷𝑶𝟒
𝟏 𝒎𝒐𝒍 𝑷𝑶𝑪𝒍𝟑
= 𝟎. 𝟔𝟐 𝒎𝒐𝒍 𝑯𝟑𝑷𝑶𝟒
Example
Step 4. Find desired units for H3PO4 Molar mass H3PO4 = 98.0 g/mol dH3PO4 = 1.83 g/mL
𝟎. 𝟔𝟐 𝒎𝒐𝒍 𝑯𝟑𝑷𝑶𝟒 ×𝟗𝟖. 𝟎 𝒈 𝑯𝟑𝑷𝑶𝟒
𝟏 𝒎𝒐𝒍 𝑯𝟑𝑷𝑶𝟒
×𝟏 𝒎𝑳 𝑯𝟑𝑷𝑶𝟒
𝟏. 𝟖𝟑 𝒈 𝑯𝟑𝑷𝑶𝟒
= 𝟑𝟑 𝒎𝑳 𝑯𝟑𝑷𝑶𝟒
Example
How many mL of C5H8 are made from 1.89x1024 molecules of C5H12?
C5H12 → C5H8 + 2H2
Step 1. Mole ratio C5H12 : C5H8 = 1:1
Step 2. Find # moles of C5H12
𝟏. 𝟖𝟗𝒙𝟏𝟎𝟐𝟒 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝟓𝑯𝟏𝟐 ×
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
𝟔. 𝟎𝟐𝟐𝒙𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝟓𝑯𝟏𝟐= 𝟑. 𝟏𝟒 𝒎𝒐𝒍𝒆𝒔 𝑪𝟓𝑯𝟏𝟐
Example
Step 3. Find # moles of C5H8
𝟑. 𝟏𝟒 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐 ×𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
= 𝟑. 𝟏𝟒 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖
Step 4. Find desired units for C5H8
Molar mass C5H8 = 68.0 g/mol dC5H8 = 0.681 g/mL
𝟑. 𝟏𝟒 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖 ×𝟔𝟖. 𝟎 𝒈 𝑪𝟓𝑯𝟖
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖
×𝟏 𝒎𝑳 𝑪𝟓𝑯𝟖
𝟎. 𝟔𝟖𝟏 𝒈 𝑪𝟓𝑯𝟖
= 𝟑𝟏𝟒 𝒎𝑳 𝑪𝟓𝑯𝟖
Practice
Given 366 mL C5H12, find the volume of C5H8 that forms from the reaction
C5H12 → C5H8 + 2H2
dC5H12 = 0.620 g/mL
dC5H8 = 0.681 g/mL
Answer: 315 mL C5H8
Practice
How many mL of C5H12 are needed to make 97.3 mL of C5H8?
C5H12 → C5H8 + 2H2
dC5H12 = 0.620 g/mL
dC5H8 = 0.681 g/mL
Answer: 113 mL C5H12
Practice
How many molecules of CO2 are produced when 79.5 g K2CO3 decompose?
K2CO3 → K2O + CO2
3.46 x 1023 molecules CO2
How many water molecules form from the combustion of 1.129 x1024 molecules C4H10?
2C4H10 + 13O2 → 8CO2 + 10H2O
5.645 x 1024 molecules H2O
Refresher
• Homework turned in and in seat at class start
• Class is dismissed when I say
• Show all work on calculations
• Do your homework (and put your name on it…)
• Ask questions in class if I’m not making sense!
• Come by in activities or after school for help!!!
Calculation Tip
Always write numbers to show:
Quantity (pay attention to sig fig!)
Units
Identity
Example: “123.0 grams NaOH”, not “123” Use “EE” function on calculator for exponents 6.022 x 1023 entered as 6.022 2nd EE 23 Displays as 6.022E23
Refresher
• Stoichiometry tells us how much chemical we need or make in a reaction
• Same process used in cooking
• I need 400 cookies. How many eggs do I have to buy?
• I have 5 pounds of flour and 1 stick of butter. How many cookies can I make?
Stoichiometry Refresher
• Coefficients in balance equation tell us how many moles of each chemical
• Ratio between coefficients called the mole ratio
• Use mole ratio to convert from amount of one chemical to amount of another
• Convert to moles (if required) before using mole ratio!!!
Example
How many mL of C5H8 are made from 1.89x1024 molecules of C5H12?
C5H12 → C5H8 + 2H2
Step 1. Mole ratio C5H12 : C5H8 = 1:1
Step 2. Convert molecules to mol C5H12
𝟏. 𝟖𝟗𝒙𝟏𝟎𝟐𝟒 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝟓𝑯𝟏𝟐 ×
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
𝟔. 𝟎𝟐𝟐𝒙𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔 𝑪𝟓𝑯𝟏𝟐
Example
Step 3. Use mole ratio to find # moles of C5H8
𝟏. 𝟖𝟗𝒙𝟏𝟎𝟐𝟒 𝒎𝒐𝒍𝒆𝒄. 𝑪𝟓𝑯𝟏𝟐 ×𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
𝟔. 𝟎𝟐𝟐𝒙𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄. 𝑪𝟓𝑯𝟏𝟐
×𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
Step 4. Find desired units for C5H8
Molar mass C5H8 = 68.0 g/mol dC5H8 = 0.681 g/mL
𝟏. 𝟖𝟗𝒙𝟏𝟎𝟐𝟒 𝒎𝒐𝒍𝒆𝒄. 𝑪𝟓𝑯𝟏𝟐 ×𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
𝟔. 𝟎𝟐𝟐𝒙𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄. 𝑪𝟓𝑯𝟏𝟐
×𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟏𝟐
×𝟔𝟖. 𝟎 𝒈 𝑪𝟓𝑯𝟖
𝟏 𝒎𝒐𝒍 𝑪𝟓𝑯𝟖
×𝟏 𝒎𝑳 𝑪𝟓𝑯𝟖
𝟎. 𝟔𝟖𝟏 𝒈 𝑪𝟓𝑯𝟖
= 𝟑𝟏𝟒 𝒎𝑳 𝑪𝟓𝑯𝟖
Limiting Reactants
Balanced chemical equation assumes exactly the right amount of each reactant
Real-world reactions don’t work this way!
Reaction goes until one reactant runs out
Limiting reactant runs out first
Excess reactant - whatever is left over
Example
Bike shop owner has 6 frames and 10 wheels and wants to build as many bikes as possible
1 frame + 2 wheels 1 bike
Builds 5 bikes with one frame left over
• Ran out of wheels “limiting reactant”
• Left over frame “excess reactant”
Example
Finding Limiting Reactant
Calculate how much product you can make from each reactant
The reactant that makes the least product is the limiting reactant
Theoretical Yield
The amount of product expected from a given amount of reactant
• Assumes perfect efficiency in the reaction
• Computed using your limiting reactant
Example
PCl3 + 3H2O → H3PO3 + 3HCl
You have 0.50 mol PCl3 and 3.0 mol H2O. Which is the limiting reactant? What is the theoretical yield of H3PO3 in grams?
0.50 mol PCl3 makes 𝟎. 𝟓𝟎 𝐦𝐨𝐥 𝐇𝟑𝐏𝐎𝟑
3.0 mol H2O makes 𝟏. 𝟎 𝐦𝐨𝐥 𝐇𝟑𝐏𝐎𝟑
PCl3 is limiting reactant and you have excess H2O
0.50 mol PCl3 makes 41 g H3PO3
Example
225 g PCl3 mix with 123 g H2O according to the reaction below. Which is the limiting reactant? What is the theoretical yield (in grams) of H3PO3?
PCl3 + 3H2O H3PO3 + 3HCl
𝟐𝟐𝟓 𝐠 𝐏𝐂𝐥𝟑 𝐦𝐚𝐤𝐞𝐬 𝟒. 𝟗𝟏 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟏𝟐𝟑 𝐠 𝐇𝟐𝐎 𝐦𝐚𝐤𝐞𝐬 𝟔. 𝟖𝟑 𝐦𝐨𝐥 𝐇𝐂𝐥
PCl3 is the limiting reactant and you have excess H2O
𝟐𝟐𝟓 𝐠 𝐏𝐂𝐥𝟑 𝐦𝐚𝐤𝐞𝐬 𝟏𝟑𝟒 𝐠 𝐇𝟑𝐏𝐎𝟑
Practice
Identify the limiting reactant and theoretical yield of HCl in grams for the reaction
PCl3 + 3H2O H3PO3 + 3HCl
Given 3.00 mol PCl3 and 3.00 mol H2O
• H2O is limiting, 1.10 x102 g HCl
Given 75.0 g PCl3 and 75.0 g H2O
• PCl3 is limiting, 59.7 g HCl
Given 1.00 mol PCl3 and 50.0 g H2O
• H2O is limiting, 101 g HCl
Actual vs. Percent Yield
Actual yield is how much product you made (actually recovered and measured)
Actual yield is always less than theoretical yield. Why?
Percent yield = 𝒂𝒄𝒕𝒖𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅× 𝟏𝟎𝟎%
Actual vs. Percent Yield
What is your percent yield if you actually collected 23.5 g of FeSO4 after an experiment that predicted a theoretical yield of 28.8 grams?
81.6%
Practice Actual vs. Percent Yield You are making ammonia via the reaction
N2 + 3H2 2NH3
If you have 57 g N2 and 4.5 g H2, calculate
a) The limiting reactant
b) The theoretical yield of NH3 in grams
c) The percent yield if you actually collected 22.7 g NH3
H2 is limiting
Theoretical yield is 26 g NH3
Percent yield is 87%
What’s left over?
How much excess reactant is left after reaction?
1) Find # mol excess reactant you started with
2) Use mole ratio between limiting reactant and excess reactant to find # mol excess reactant you used
3) Subtract “used” from “started with”
Example
225 g PCl3 mix with 123 g H2O according to the reaction below. How much H2O is left over?
PCl3 + 3H2O H3PO3 + 3HCl
1) Find # mol excess reactant you started with
6.83 mol H2O
2) Use mole ratio between limiting reactant and excess reactant to find # mol excess reactant you used
4.91 mol H2O used
3) Subtract “used” from “started with”
1.92 mol (34.6 g) H2O remaining