topic 3 - quantitative chemistry Flashcards
relative formula mass, balancing equations, avogadro's constant, moles, conservation of mass, reacting masses, limiting reactants, % mass, % yield, atom economy
how do you calculate relative formula mass
the sum of the atomic mass number of all the atoms in the molecular formula added together
calculate the relative formula mass of these:
1) H₂O
2) CaCO₃
3) Al₂(SO₄)₃
1) H + H + O
-> 1 + 1 + 16 = 18
2) Ca + C + O + O + O
-> 40 + 12 + 16 + 16 + 16 = 100
3) Al + Al + S + S + S + O + O + O + O + O + O + O + O + O + O + O + O
-> 27 + 27 + 32 + 32 + 32 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 + 16 = 342
how do you balance equations
1) write out the amount of atoms on the reactants and products side
2) see what is different and put big numbers in front of the elements/compounds that need to change to become equal to their corresponding element on the other side
3) do this until both sides are equal
e.g.
H₂ + O₂ –> H₂O
H - 2 —–> H - 2
O - 2 —–> O - 1 - not equal
H - 2 —–> H - 2
O - 2 —–> O - 2 - x2 to make it equal
H₂ + O₂ –> 2H₂O - have to add in the changes
H - 2 —–> H - 4 - because 2x2 is 4
O - 2 —–> O - 2
H - 4 —–> H - 4 - x2 to make it equal
O - 2 —–> O - 2
2H₂ + O₂ –> 2H₂O - have to add in the changes
done //
what is avogadro’s constant
6.02 × 10²³ –> 1 mole of any element or compound contains this many atoms or molecules
1 mole of any substance in grams is equal to its RFM in grams
H₂O RFM = 18 –> 1 mole = 18 grams
moles equation
number of moles = mass/RFM
mass = number of moles x RFM
the number in front of an element/compound affects the mass
what is the conservation of mass
the total of all the reactants is equal to the total mass of all the products in a chemical reaction
e.g.
CaCO₃ —> CaO + CO₂
200g —–> 112g + ???
200 - 112 = 88g
SO!!
200g —–> 112g + 88g
reacting masses
if you know the mass of a reactant and a product, you can use simple ratios to calculate reacting masses and product masses
e.g.
when 12 g of carbon is burned in the air, 44 g of carbon dioxide is produced. what mass of carbon is needed to produce 11 g of carbon dioxide?
12 g of carbon = 44 g of carbon dioxide
12 ÷ 44 g = 1 g of carbon dioxide
for 11 g of carbon dioxide:
11 × (12 ÷ 44 g) = 3 g of carbon
OR
12 g of carbon = 44g of carbon dioxide
?? g of carbon = 11g of carbon dioxide
44 –> 11 = divide by 4
12 –> ?? = divide by 4
12/4 = 3 g of carbon
limiting reactants
the reactant that is all used up is the
limiting reactant - it sets a limit on how much product is produced
the reactant that is left over is in excess
no more product can form when the limiting reactant is all used up
how to find the number of particles or molecules in any particle or molecule
no. of particles/molecules
= no. of moles x 6.02 x 10²³
how to find the mass percent of an element or compound
mass of element/
total mass of compound x 100
how to find the concentration of a solution with moles and volume
concentration (mol/dm³)
= moles (mol)/volume of solution (dm³)
how to find the concentration of a solution with mass and volume
concentration(g/l) = mass (g)/volume(l)
percentage yield equation
actual mass/theoretical mass x 100
what is percentage yield
it shows how much product was actually made compared with the amount of product that was expected
theoretical yield - maximum amount of product expected (calculate)
actual yield - amount of product that is actually obtained from the real chemical reaction (told this)
atom economy equation
atom economy
= desired product/total RFM of all reactants x 100
