There are many devices & units that are used to quantify properties of matter
All measuring devices are limited in their accuracy
To communicate the accuracy of a measurement, we used significant figures (or digits)
All counted values are significant (in essence, meaningful) figures (or sig fig for short)
Significant figures, when using a measuring device, are all of the values that can be read off of the device, as well as one that is estimated
Even with electronic measurement devices, the last digit is an estimate
A and P Method:
Cannot be more accurate than the least accurate measurement
When adding or subtracting values, the answer will have the same number of digits after the decimal as the least accurate measurement
Chop off what is more accurate than the least accurate measurement
Chop it off, you can't calculate a number you don't have!!!
The sig fig in the answer from multiplying and dividing measured values have the same number of total sig figs as the value with the least amount of sig figs in the question
Chop off digits that have more than the least amount of sig figs in the question
Write the name or unit of the quantity that you're looking for & an equal sign; this is your target
Write the given value that you have to begin with
Multiply the given value by conversion factors arranged so that the given units are cancelled (divided out) and the target unit remains
Benefits of using this method:
Quantities of invisible tiny particles Quantities of measurable amounts
like molecules and atoms ⟺ like mass and volume
If you know the relative mass of 2 items, then you can always determine when you have equal quantities of both
For compounds: just ^^add the mass of each element in the compound together^^
Eg. what is the mass of 15.5 moles of carbon dioxide (C = 12, 0 = 16 (x2) – 44 g)
Mass = 15.5 mol = 44 g = 682g Use factor label method!
1 mol
@@6.02 x 1023 particles/mol@@
Note: not all substances are made of molecules! The particles that make up most elements are atoms. Ionic compounds are a repeating pattern of ions, the smallest of which can be called a formula unit.
Eg. One mole of water (H2O -- 18g/mol) contains 2 moles of Hydrogen (1g/mol + 1g/mol = 2g/mol) and 1 mole of Oxygen (16g/mol)
Grams of element you're looking for x 100% = % of element you were looking for
Grams of the compound you have
MASS → MOLES → LOWEST WHOLE NUMBER RATIO
Eg. a sample was found to contain 50% sulfur and 50% oxygen by mass - what is the empirical formula?
Then, you can see that the formula created is SO2 because the lowest wholes represent the quantities of the elements in the formula
Sometimes…the moles of each element do not give you a nice even whole number, and you'll need to multiply to find the best ratio
For a ratio of 1:1.5 ⇒ multiply by 2 to get 2:3
For a ratio of 1:1.333 ⇒ multiply by 3 to get 3:4
For a ratio of 1:1.25 ⇒ multiply by 4 to get 4:5
Sometimes, the percentages are not given - masses need to be found in another way….
Eg. a 50.51 g sample of a compound made from iron and oxygen is decomposed, and 35.36 g of the iron remains. Find the empirical formula
Mass Moles Lowest whole
Fe: 35.36 x 1 mol Fe = 0.6331 mol ⇒ 1 x 2
55.85 g 0.6331
O: 50.51-35.36 = 15.15 x 1 mol O = 0.9469 mol ⇒ 1.5 x2
(subtract whole sample 16 g 0.6331
from Fe which was also
given)
Eg. If the empirical formula of a molecule is CH3 and the molar mass of the compound is 30 g/mol, what is the molecular formula?
Eg.
2H2 + O2 → 2H2O
2 particles H2 + 1 particle O2 → 2 particles H2O
2 moles H2 + 1 mole O2 → 2 moles H2O
Eg. If 5.0 moles of Hydrogen are consumed, how many moles of oxygen are also consumed?
Moles O2 = 5.0 mol H2 x 1 mol O2 = 2.5 mol O2
2 mol H2
Eg. if 5.0 moles of oxygen are consumed, how many moles of water are produced?
Moles H2O = 5.0 mol O2 x 2 mol H2O = 10.0 mol H2O
1 mol O2
Use the quantities from the balanced equation (equivalent in moles) to calculate what you need
Stoichiometry: involves 3 steps to determine the quantity involved in a chemical reaction
Eg. Quantity consumed: what mass of oxygen is consumed when reacting with 5.0 g of Hydrogen?
2H2 + O2 → 2H2O
Step 1 - convert into moles
Moles H2 = 5.0 g H2 x 1 mol H2 = 2.5 mol H2
2g
Step 2 - convert the given moles to the moles of the quantity you want (target)
Moles O2 = 2.5 mol H2 x 1 mol O2 = 1.25 mol H2
2 mol H2
Step 3 - convert target moles to the unit you want
Mass O2 = 1.25 mol O2, x 32g = 40 g O2
1 mol
Putting it all together:
Mass O2 = 5.0 H2 x 1 mol H2 x 1 mol O2 x 32g = 40 g O2
2g 2 mol H2 1 mol O2
[target unit] [target element/compound] = [given amount of given e/c] x [1 mol/molar mass of given e/c] x [quantity of target e/c in balanced equation/quantity of given e/c in balanced equation] x [conversion factor to get into target units]
A chemical reaction is like a factory with input components (reactants) and output (products)
If one of the reactants runs out, the reaction will then stop
Therefore, that reactant then controls how much product can be made
The reactant that runs out = the limiting reagent
The leftover reactant = the excess reagent
Whenever you have more than one quantity of a reactant, you will need to determine the limiting reagent before any other quantities can be found
The smaller value is the limiting reagent
Eg) if 3.0g of H2 reacts with 6.0g of O2
2H2 + O2 → 2H2O
H2 : O2 :
3.0g H2 x 1 mol = 1.5mol H = 0.75 6.0g O2 x 1 mol = 0.25 = 0.25
2g 2 32g 1
0.25 is smaller than 0.75 ∴ O2 is the Limiting Reagent
Yield = the quantity of a product that can be expected to be made
Eg) What mass of water can be formed from 3.0g of H2 reacting with 6.0g of O2?
*********you need to use the limiting reagent to solve this question!!!
MassH2O = 6.0g O2 x 1 mol O2 x 2 mol H2O x 18g H2O = 6.8g H2O
(yield) ↑ 32 g 1 mol O2 1 mol
Given quantity LR ↑ ↑ ↑
1 mol/molar mass target/given target molar mass
of LR equation to get into grams
coefficients
Excess Reagent: what mass of excess reactant remains from 3.0g of H2 reacting with 6.0g of O2?
*******you also need to use the LR here too
MassH2 = 6.0g O2 x 1 mol O2 x 2 mol H2 x 2gH2 = 0.75g H2
(used) ↑ 32g 1 mol O2 1 mol
Given quantity LR ↑ ↑ ↑
1 mol/molar mass target/given target molar mass
of LR coefficients to get into g
Remaining H2 = 3.00g - 0.75g = 2.25g
↑ ↑
Given Used
amount amount
Percentage yield = Actual yield x 100%
Theoretical yield
Percentage purity = Mass of a pure substance x 100%
Total mass of a sample
Percent error = |Experimental value - Accepted value| x 100%
Accepted value