topic 9 calculations involving mass

Question 1

how do you calculate the relative formula mass of a substance?

do it for alluminium chloride - AlCl₃

²⁷₁₃Al, ^35.5₁₇Cl

Answer 1

add together the RAM of each atom in 1 molecules of the substance:

27 + (35.5 * 3) = 133.5

Question 2

how do you calculate the empirical formula of a substance?

try: calculate the empirical formula of a compound made by combining 1.92g of magnesium with 5.68g of chlorine

Answer 2

divide the mass of each element by its RAM:

Mg > 1.92/24 = 0.08, Cl > 5.68/35.5 = 0.16

divide both amounts by the smaller:

Mg > 0.08/0.08 = 1, CL > 0.16/0.08 = 2

ratio of elements = 1Mg:2Cl

empirical formula = Mg₁Cl₂ or MgCl₂

Question 3

what is the difference between the empirical formula and moleclar formula of a substance?

Answer 3

the empirical is the simplest ratio of elements in a substance, the molecular formula is the actual formula of a substances molecules

Question 4

what is the law of conservation of mass in a closed and non-enclosed system?

Answer 4

closed: mass cannot be destroyed or lost,

non-enclosed: any “lost mass”would be from any gass being created not being weighed

Question 5

what is meant by 1 mole?

Answer 5

6*10²³ particles

Question 6

what is the equation of molex, mass and RAM?

Answer 6

Moles = Mass/RAM

Question 7

describe an experiment to find the empirical formu;a of a compound(magnesium oxide)

Answer 7

1. heat a crucible untill its red hor to rewmove any substance from a previous experiment
2. leave to cool then weigh the crucible and its lid
3. add a ribbon of magnesium and reweigh
4. heat over a bunsen burner untill the magnesium is white(have the lid on but leave a small gap, so none escapes but oxygen can get in
5. find the weight of the magnesium and oxygen(subtract the initial measurment from the second to get the magnesium mass than subtract the mass of the second measurement from the third to get the mass of oxygen then put through the steps to work out the empirical formula

Question 8

how do you use “easy moles R magic”?

use it to work outhow much aluminium oxide would be needed to make 4.5g of alluminium

2Al₂O₃ -> 4Al + 3O₂

Answer 8

1. equation: 2Al₂O_<u>3</u> -> 4__Al + 3O₂ (underline or circle the 2 substances we are asked about)
2. moles: 4.5/27=0.16(mass/RAM of aluminium to work out how many moles are needed)
3. ratio: 2:1(the 2 infront of aluminium oxide shows there should be 2 moles for every 4 of aluminium, so we do 0.16 / 2 = 0.08 as there are twice as many moles of aluminium)
4. mass: 0.08 * 102(RFM of aluminium oxide) = 8.16g