Module 2 Yr 11 Flashcards
State the law of conservation of mass
Matter cannot be created or destroyed only rearranged meaning total mass of reactants must be equivalent to the total mass of the products.
State the law of definite proportions
A certain compound always contains the same elements in the same proportions by mass
Pure water (H2O) will always be 88.8% oxygen and 11.2% hydrogen by mass
Define stoichiometric coefficients and stoichiometric ratio
numbers in front of chemicals whilst balancing chemical equations
AKA molar ratio: ratio between these stoichiometric coefficients, is independent of mass and depends only on amount of substance
State Gay-Lussac’s law of combining volume
Volume ratio of the gases is equal to their stoichiometric ratio
2H2(g) + O2(g) → 2H2O (g)
2L: 1L → 2L
Note: law only applies for gases only
Example 2H2(g) + O2(g) → 2H2O(l) doesn’t work for this law
How to find mass ratio
Example: In a certain reaction, hydrogen gas reacts with oxygen gas to form water. Find the mass ratios of each of the reactants and products required for the reaction. In this example, take the mass of an oxygen atom to be 16u and the mass of a hydrogen atom to be 1u.
- Balance chemical equation
- Count the number of atoms of each element in each species
- Multiply this number by the stoichiometric coefficient
- For each element, multiply the number of atoms by its atomic mass
- Add up the mass of each element in the species to find its total mass in u
- Find the simplest ratio between masses of species
Ans example
1. 2H2 (g) + O2 (g) → 2H2O (g)
2. 4H 2O. 4H+2O
3. 4x1u. 2x16u. 4x1u+2x16u
Mass ratio: 1:8:9
Define mole
The number of atoms in exactly 12g of carbon-12. Approximately 6.022 X 10^23 (Avogadro’s number NA)
Size of relativity between atomic mass unit and gram
A mole of thins each weighing 1 unified atomic mass unit will weigh 1g in total
6.022 X 10^23u = 1g
Number of moles, Avogadro’s number and number of particles formula
n= N/ NA
n: number of moles (mol)
N: number of particles (atoms, formula units, molecules)
NA: Avogadro’s number
Units for ionic or covalent networks and metals
Formula units not molecules
Formula to calculate molar mass formula
n=m/MM
n: number of moles (mol)
m: mass (g)
MM: molar mass (gmol^-1)
Relationship between stochiometric ratio and molar ratio in chemical equation
Both are equal
How to calculate percentage composition?
Calculate percentage composition of constituent elements of iron (III) oxide
- Calculate molar mass of compound
- Find mass of desired element in 1 mole of compound
- Divide number by molar mass of the compound
Ans example:
mm(Fe2O3)=2 x 55.85 + 3 x 16=159.7 g mol^-1
mm(Fe2)= 2 x 55.85=111.7 g mol^-1
%Fe=mm(Fe2)/mm(Fe2O3) x 100=111.7/159.7 x 100=69.94% (4sf)
mm(O3)=3 x 16=48 g mol^-1
%O=mm(O3)/mm(Fe2O3) x 100=48/159.7 x 100=30.06% (4sf)
Define empirical formula and molecular formula
Simplest whole number ratio of elements present in a compound. Often used for ionic or covalent network compounds
The exact number and type of atoms present in a molecule. Used for covalent molecular structures
Example: dinitrogen tetroxide molecular formula is N2O4
Empirical formula is NO2
How to calculate empirical formula HAVENT FINISHED BANNED ACCESS
Example: Determine molecular formula and empirical formula of succinic acid if it is 40.7% carbon, 5.1% hydrogen and 54.2% oxygen by mass. Molar mass is 118.09g mol^-1.
- Assume mass of compound is 100g and calculate the constituent mass of each element
- Determine the number of moles of each element
- Determine the simplest molar ratio between elements by setting lowest value to be 1
- Write down empirical formula
To find molecular formula, do above steps and then determine what multiplying factor needs to be applied to satisfy MM of compound (will be given)
Ans Example: C H O
Assuming mass=100. 40.7 5.1. 54.2
Calculate n n=40.7/12.01 n=5.1/1.008 n=54.2/16
n=3.39 n=5.06 n=3.39
Divide by smallest n. 1 1.4……. 2
Integer ratio. 2 3 2
Empirical Formula: C2H3O2
mm(C2xH3x)2x)=118.09g=x MM(C2HO3)
mm(C2)=12.01 x 2=24.02 g mol^-1 (4sf)
mm(H3)=1.008 x 3=3.024 g mol^-1 (4sf)
mm(O2)=16 x 2 =32g mol^-1
mm(C2H3O2)=12.01 x 2 +1.008 x 3 +16 x 2=59.044
118.09/59.044=2
Molecular formula: C4H6O4
How to solve limiting reagent calculations
Example: 11.17g of iron reacts with 16g of oxygen gas to form iron (III) oxide. Find mass of iron (III) oxide formed.
- Construct balanced chemical equation
- Determine number of moles of all reagents
- Based on molar ratios, determine which reagent is the limiting reagent
- Use the number of moles of limiting reagent to calculate the number of moles of product
Ans Example:
1. 4Fe (s) + 3O2 (g) → 2Fe2O3 (s)
2. n=m(Fe)/mm(Fe) n=m(O2)
n=11.17/55.85 n=16/32
n=0.2 n=0.5
3. n(Fe)=0.2. 4:3 3:4
n(Fe):n(O2) n(O2) : n(Fe)
0.2:0.5 0.5: 0.67
4. n(Fe2O3)=1/2 x n(Fe)
=0.1 mol
5. mm(Fe2O3)=(55.85 x 2)+(16 x 3)=159.7g mol^-1
m(Fe2O3)=mm(Fe2O3) x n(Fe2O3)
=0.1 x 159.7
=160g
Define concentration
Amount of solute dissolved per unit of solvent
Identify types of measures of concentration
Percentage weight by weight %(w/w): equal to g/100g
Mass of solute as percentage of total mass of solution
Percentage weight by volume %(w/v): equal to g/100mL
Mass of solute as percentage of total volume of solution
Percentage volume by volume %(v/v): equal to mL/100mL
Volume of solute as percentage of total volume of solution
Parts per million (ppm): equal to mg/kg or µL/L
Mass of volume of solution per million units of mass or volume
Molarity (M): equal to mol/L
No of moles of solute per litre
Define molarity
State formula
Number of moles of solute per litre of solution.
c=n/V
c: concentration (mol/L)
n: number of moles (mol)
V: volume (L)
Square bracket notation to denote its molarity
[NaCl]: molarity of NaCl
If asked for concentration of substance, molarity is implied
Define dilution
Process to lower concentration of a substance by adding more solvent but keeping amount of solute constant
How to calculate new concentration following dilution (method 1)
Using Method 1, calculate the concentration of HCl in final solution when 10.0L is added to a 5.0L solution of 10.0 mol L^-1 HCl
Method 1: Find no of moles first
1. Find no of moles of solute
2. Find new total volume
3. Use c=n/V with new volume
Ans example:
1. n(HCl)=[HCL] x V(HCl) = 10 x 5 =50mol
2. Find new volume=5 + 10=15L
3. [HCL]=n(HCl)/V(HCl)=50/15=3.3M (2sf)
How to calculate new concentration following dilution (method 2)
Using Method 2, calculate concentration of HCl in final solution when 10.0L of water is added to a 5.0L solution of 10.0 mol L^-1 HCl
n1=n2
n1: no of moles of solute prior to dilution
n2: no of moles of solute after dilution
c1V1=c2V2
c1: concentration prior to dilution
V1: volume prior to dilution
c2: concentration after dilution
V2: volume after dilution
Ans Example:
c1V1=c2V2
10 x 5=c2 x 15
c2=50/15
c2=3.3M (2sf)
Name of chemical formula and charge of polyatomic ions
Nitrate
Hydroxide
Hydrogen carbonate/bicarbonate
Carbonate
Sulfate
Phosphate
Ammonium
Chemical Formula Charge
NO3^- -1
OH^- -1
HCO3^- -1
CO3^2- -2
SO4^2- -2
PO4^3- -3
NH4^+ 1
When NaOH (aq) is reacted with H2SO4 (aq), the following reaction occurs: 2 NaOH (aq) → Na2SO4 (aq) + 2H2O (l)
Find the mass of Na2SO4 (aq) produced if 100.0 mL of 0.500 mol L^-1 sodium hydroxide reacted with 200.0 mL of 0.220 mol L^-1 sulfuric acid.
CHECK with video
1.n(NaOH)=[NaOH] x V(NaOH) n=(H2SO4)=[H2SO4] x V(H2SO4)
=0.5 x 0.1 =0.2 x 0.22
=0.05 mol =0.044 mol
2. Finding limiting reagent 3.n(Na2SO4)=1/2 n(NaOH)
Assuming all of H2SO4 is used =1/2 x 0.05
n(NaOH)=0.088 =0.025 mol
2 x n(H2SO4) <n(NaOH) 4.m(Na2SO4)=n(Na2SO4)xmm(Na2So4)
NaOH limiting =0.025 x (22.99x2)+(16x4)+32.07
=3.55g
Define standard solution
Define Primary standard solution
A solution that has a known fixed concentration.
A standard solution made by dissolving a known amount of solute in known amount of solvent. Only certain solutes can be used.
