calculations

Question 1

isotopes

Answer 1

• atoms of the same element which have the same number of protons but a different number of neutrons
• same atomic number but a different mass number
• same number of electrons
• They react identically to atoms of the same element as the both have the same number of electrons

Question 2

relative atomic mass (Ar) and relative formula mass(Mr)

Answer 2

the Ar of an element is the same as its mass number
The Mr of a compound is the same as all the Ar s in that compound
e.g CaCO3 : Ca C O3 49 12 + (3*16) = 100

Question 3

the mole

Answer 3

the Ar or Mr of a substance in grams is known as one mole of that substanceFe has an Ar of 56. so one mole of iron will weigh 56 grams.mass= moles*Mr
Mr=mass/moles
moles= mass/mr

Question 4

percentage by mass

Answer 4

%mass = (Ar * number of atoms of that element/)
(Mr of compound) 100
E.G % mass of beryllium in BeO
(91 / 9+16) * 100 = 36% Ar= just Be Mr=BeO

Question 5

empirical formula

Answer 5

simplest ratio of atoms of each atoms of each element in a compound

1) list all the elements in the compound side by side
2) underneath them write their experimental masses or percentages as given in the question
3) divide each mass or percentage by the Ar of that particular element(top number)
4) divide each answer in step 3 by the lowest of all the answers
5) the answer should be the simplest ratio

Question 6

molecular formula

Answer 6

actual number of atoms of each element in the compound
the empirical formula is used for ionic compounds, but for covalent bonds it is not alway the same as the molecular formular
e.g. molecular formula of magnesium chloride (ionic): MgCl2
empirical formula: MGCl2
molecular formula of ethane(covalent): C2H6
empirical formula: CH3

Question 7

percentage yields and reversible reactions

Answer 7

compares predicted yield with the actual yield
the more reactants you start with the larger the actual yield
%yield= (actual yield/predicted yield)*100
YIELD IS NEVER 100%