1) since we are given percentages, we can assume we have 100 grams of the molecule.
55.6 % Cu ----> 55.6 grams Cu
16.4 % Fe------> 16.4 grams Fe
28.0% S--------> 28.0 grams S
2) convert each gram to moles using the molar masses given



3) we divide the smallest value of moles (0.293) to each one.
Cu --> 0.876 / 0.293= 3
Fe---> 0.293 / 0.293= 1
S-----> 0.875 / 0.293= 3
4) let's write the empirical formula
Cu₃FeS₃
Answer:
2 half-lives=0.8
6 half-lives= 0.05
Explanation
Half-lives are constant and always decrease by half, implying that the concentration decreases by half at a consistent rate.
3.2/2= 1.6/2= 0.8 is two half-lives
3.2/2= 1.6/2= 0.8/2= 0.4/2= 0.2/2= 0.1/2=0.05 is six half-lives
According to the stoichiometry the CO2 formed is 1/2 the moles of H2O. So take the moles of CO2 as 5moles.
n=m/M
Where :
n = mole
m = mass
M = molar mass
Molar mass of CO2 = 44gmol^-1
Therefore mass = 5mol× 44gmol^-1= 220g