A) Picture D
B) Chlorine
C) Natrium chloride (NaCl) / salt
Answer:
C₂H₃Cl₃O₂
Explanation:
Given parameters:
Percentage composition:
14.52% C
1.83% H
64.30% Cl
19.35% O
Molar mass = 165.4 g/mo
Unknown:
Empirical formula = ?
Solution:
The empirical formula of a compound is the simplest formula of a compound. To solve this problem, let us follow the process below:
C H Cl O
%composition 14.52 1.83 64.3 19.35
Molar mass 12 1 17 16
number of
moles 14.52/12 1.83/1 64.3/3.5 19.35/16
1.21 1.83 1.81 1.21
Divide by
the smallest 1.21/1.21 1.83/1.21 1.81/1.21 1.21/1.21
1 1.51 1.51 1
multiply
by 2 2 3 3 2
Empirical formula C₂H₃Cl₃O₂
1) We need to convert 12.0 g of H2 into moles of H2, and <span> 74.5 grams of CO into moles of CO
</span><span>Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol
</span>12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO
<span>2) Now we can use reaction to find out what substance will react completely, and what will be leftover.
CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)
How much
we need CO? 3 mol 6 mol
We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.
3) </span>CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol
4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol
2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH
<span>
Answer is </span>D) 85.12 grams.
