Moles, Stoichiometry, Limiting and Excess Reactants and Balancing Chemical Equations.

Question 1

What number is a mole/avagadro’s number?

Answer 1

6.022 x 10^23

Question 2

A mole of an element is proportionate to it’s:

Answer 2

Average atomic mass. Example: Carbon, 1 mole of carbon is equal to 6g of carbon

Question 3

How do you calculate molar mass?

Answer 3

Molar mass is basically the total mass in amu, remove the amu and add g/mol. Example: Oxygen’s molar mass us 16.01 g/mol. Ozone (O3) molar mass is 48.03 g/mol

Question 4

How do you convert from grams to moles in single atoms?

Answer 4

G/Molar mass

Question 5

How do you convert from grams to moles if there you’re only trying to find the number of moles of one atom in a compound with many atoms?

Answer 5

(G of compound)/(Molar mass of compound)*(Number of moles of the atom)

Question 6

How do you convert from moles to grams?

Answer 6

G = (N. of Moles) * (Molar mass of substance you are trying to find grams of)

Question 7

How do you convert moles to atoms in atoms?

Answer 7

Atoms = (N. of moles) x (6.022 * 10^23/Avagadro’s number)

Question 8

How do you convert from moles to atoms in molecules?

Answer 8

Atoms of Molecules = (N. of mols) * (6.022 * 10^23) * (N. of atoms per molecule).

Question 9

How do you convert atoms to moles?

Answer 9

​Mols = (N. of atoms) / (6.022 * 10^23)

Question 10

How do you convert from grams to atoms/molecules?

Answer 10

Atoms/Molecules = (N. of grams) / (Molar mass of atom/molecule) x (6.022 x 10^23

Question 11

How do you convert from atoms/molecules to grams?

Answer 11

Grams = (N. of atoms) / (6.022 x 10^23) x (Molar mass of substance)

Question 12

How do you convert from atoms to grams if you’re only trying to find the number of a specific thing?

Answer 12

Grams = (N. of atoms) / (6.022 x 10^23) / (The amount less) x (Molar mass of the substance)

Egs: P4 from N. of atoms phosphorus. Amount less = 4

Question 13

How do you balance a chemical equation?

Answer 13

Calculate the number of each element on BOTH sides of the equation. If one side has less of an element, you add coefficients to that element (and other ones if they aren’t multiples). You also need to make sure you do not change the subscript! (Example: Mg + N2 = Mg3N2 is not balanced because Mg only has one and Mg3N2 has 3 magnesiums. The correct way of writing it is 3Mg + N2 = Mg3N2)

Question 14

When balancing chemical equations, when happens when you need a fraction coefficient?

Answer 14

Multiply all coefficients by the denominator.

Question 15

In stoichiometry, what are four types of molar problems?

Answer 15

Mol A –> Mol B
Mol A –> Gram A
Gram A –> Mol A
Gram A –> Gram B

Question 16

How do you solve Mol A to Mol B stoichiometry problems?

Answer 16

Assume we know the number of moles in mol A called “A”. First, write a chemical equation of Mol A to Mol B. Than, calculate the molar ratio between Mol A and Mol B, which would just be (Coefficient of Mol B) / (Coefficient of Mol A). Multiply A by the molar ratio. Example: 3.4 moles of SO2 reacts with oxygen gas to make SO3. This would be 2SO2 + O2 = 2SO3 which would be 3.4 x 2/2 = 3.4 moles of mol B

Question 17

How do you mol A to gram B stoichiometry problems?

Answer 17

Assume we know the number of moles in mol A called “A”. First, write a chemical equation of Mol A to Mol B. Than, calculate the molar ratio between Mol A and Mol B, which would just be (Coefficient of Mol B) / (Coefficient of Mol A). Then calculate the molar mass of gram B and multiply A by the molar ratio and the molar mass of gram B.

Question 18

How do you solve Gram A to Mol B stoichiometry problems?

Answer 18

Assume we know the number of moles in mol A called “A”. First, write a chemical equation of Mol A to Gram B. Than, calculate the molar ratio between Mol A and Gram B, which would just be (Coefficient of Mol B) / (Coefficient of Mol A). Then calculate the molar mass of gram B. The end result is (A) / (Molar mass of B) * (Molar ratio).

Question 19

How do we solve gram A to gram B stoichiometry problems?

Answer 19

Assume we know the number of grams in Gram A called “A”. First, write a chemical equation of Gram A to gram B. Than, calculate the molar ratio between Gram A and Gram B, which would just be (Coefficient of Mol B) / (Coefficient of Mol A). Then calculate the molar mass of gram A and Gram B. The end result is (A) / (Molar mass of gram A) * (Molar Ratio) * (Molar mass of gram B)

Question 20

What is a limiting reactant?

Answer 20

A limiting reactant is the element or compound that runs out first in a chemical equation.

Question 21

How do you find the limiting reactant in a chemical equation knowing the number of atoms/molecules?

Answer 21

First, balance the chemical equation. On the left side of the equation, calculate each atoms/molecules atom:coefficient ratio. Whichever has the lowest is the limiting reactant. This method also works given the number of moles.

By lowest, we mean by division. Example: 4:2 is actually 4/2 so it’s 2.

Question 22

How do you find the limiting reactant in a chemical equation knowing the number of grams?

Answer 22

First, balance the chemical equation. Next, convert grams to moles. Then, find the mole:coefficient. Whichever has the lowest mole:coefficient ratio is the limiting reactant.

By lowest, we mean by division. Example: 4:2 is actually 4/2 so it’s 2.

Question 23

If you’re given the number of moles of each reactant in a chemical equation, how can you find the amount of a product in moles?

Answer 23

First, balance the chemical equation. Then calculate the following for each reactant: (Moles of reactant) x (Coefficient of product/Coefficient of reactant). Whichever one produces the lowest amount of product is the correct answer (since it’s the limiting reactant).

Question 24

What is theoretical yield?

Answer 24

Maximum amount of product you can get with the limiting reactant.

Question 25

If you’re given the number of grams of each reactant in a chemical equation, how can you find the amount of a product in grams?

Answer 25

(N. of grams in reactant) / (Molar mass of reactant) x (Coefficient of product/Coefficient of reactant) x (Molar mass of product).
Repeat this for all reactants.
The one that’s lowest is correct.

Question 26

Given the grams of each reactant in a chemical equation, identify how much of the excess reactant is consumed in grams.

Answer 26

First balance the chemical equation and find the Limiting reactant (LR). The other is the excess reactant (ER).
The final result:
(G. of LR) / (Molar mass of LR) x ((Coefficient of ER) / (Coefficient of LR)) x (Molar mass of ER)

Molar mass is for conversions to grams. Uneccecary for mole or atoms.

Question 27

What is the difference between theoretical yield and actual yield?

Answer 27

Theoretical yield is the MAXIMUM amount of product you can get.
Actual yield is how much product you get from actually doing the experiment.

Question 28

What is percent yield?

Answer 28

Percent yield = (Actual Yield) / (Theoretical yield) x 100%

Question 29

How do you calculate percent error?

Answer 29

Percent error = Abs( (Measured - Actual) / (Actual) )

Question 30

How do you calculate the average atomic mass given the mass in amu and the relative abundance of every isotope?

Answer 30

Average = M1P1 + M2P2 … (if there is more)

Question 31

How do you calculate the relative abundance if given the mass of two isotopes and the mass of the element?

Answer 31

Average = M1P1 + M2P2
(Plug) = (Plug)(x) + (Plug)(1-x)
(Plug) = (Plug)(x) + (Plug) - (Plug)(x)

X should be a decimal representing a percentage