CH201 Homework 1(c): Stochiometry Please show all calculations and description s
ID: 1022838 • Letter: C
Question
CH201 Homework 1(c): Stochiometry
Please show all calculations and description so I can clearly understand stochiometry, as I missed first few class lectures.
A. What is the formula of a compound if a sample of the compound contains 0.211 mol X, 0.317 mol Y, and 0.422 mol Z?
B. What is the simplest formula of a rhenium oxide that is 82.32 % Re by mass? Write the symbol of the metal first.
C. What is the empirical formula of a compound that is 56.95% Cu, 14.37% S, and 28.68% O? Write the elements in the order Cu, S, and O.
Explanation / Answer
A] Given 0.211 mol X, 0.317 mol Y, and 0.422 mol Z
Divide by small number ==>
X = 0.211 / 0.211 = 1
Y = 0.317 / 0.211 = 1.5
Z = 0.422 / 0.211 = 2
Make them to near integers ;
So multiply by 2
the formula is X2Y3Z4
B] Begin by assuming that you have 100 grams of the metal oxide. In that case there will be 82.32 grams of Re and 17.68 g of oxygen.
Then covert the masses to moles.
82.32 g Re x (1 mol Re / 186.2 g Re) = 0.4421 mol Re
17.68 g O x (1 mol O / 16.0 g O) = 1.105 mol O
0.4421/0.4421 = 1
1.105 / 0.4421 = 2.5
Divide both by the smaller number of moles and get
1 mol Re and 2.5 mol O
Multiply by 2 then it is
The formula is Re2O5
C] Suppose 100gms of compound
Moles of Cu = 0.896
Moles of S = 0.449
Moles of O = 1.7925
divide all by smallest number of moles
Cu = 2
S =1
O = 4
THerefore empirical formula is Cu2SO4