Question

15. The percent composition by mass of a compound is 76.0% C, 12.8% H, and 11.2%0. The molar mass of this compound is 284.5 g/mol. What is the molecular formula of the compound? A) C1oHO B) C9H10 C) C16H30, D) C20H30, E) C13H4O2 16. The percent composition by mass of an unknown chlorinated hydrocarbon was found to be 37.83% , 6.35% H, and 55.83% Cl by mass. What is the empirical formula of this compound? A) CH.CIB) CH,CIC) CH.Cl? D) CH.CIE) CHCI 17. An oxyhydrocarbon produced in a chemical reaction was determined to have a molecular mass of 120.20 g/mol. Which of the following is a possible empirical formula for this compound? A) CH40 B) C2H6O C) C3H8O D) CHỤO E) C4H10O4 18. An organic thiol compound is 38.66% C, 9.73% H, and 51.61% S by mass. What is the empirical formula of this compound? A) CH.S B) CHS C) C. HSD) CHISE) CsHS 19. What is the coefficient of H2O when the following equation is properly balanced with the smallest set of whole numbers? Na + H20 NaOH + H2 A) 1 B) 2 C) 3 D) 4 E) 5

Answer 1

15.

Firstly assuming there is 100 g of compound so the amounts become

C= 76% = 76 g

H=12.8% = 12.8 g

O=11.2% =11.2 g

Now calculating the no. of moles (No. of moles = )

Given mass Molar mass

No. of moles of Carbon =    = 6.33 mole of C

No. of moles of Hydrogen = = 12.7 mole of H

Mo. of moles of Oxygen = = o.7 mole of O

Now for empirical formula dividing all the moles by the smallest mole value and that is of Oxygen = 0.7

Carbon = = 9

Hydrogen =    = 18

Oxygen =   = 1

So empirical formula will be C ₉ H ₁₈ O

Now Molecular formula = n × empirical formula

And n =  

Mass smperical mass

n =      = 2                                (empirical mass of C ₉ H ₁₈ O = 9×12+1×18+16×1 = 142)

Molecular formula = 2 × C ₉ H ₁₈ O

               Molecular formula    = C₁₈H₃₆O₂

So option (E) C₁₈H₃₆O₂ is correct

16.

Firstly assuming there is 100 g of compound so the amounts become

C= 37.83% = 37.83 g

H=6.35% = 6.35 g

Cl=55.83% =55.83 g

Now calculating the no. of moles (No. of moles = )

Given mass Molar mass

No. of moles of Carbon =    = 3.1 mole of C

No. of moles of Hydrogen = = 6.3 mole of H

Mo. of moles of Chlorine = = 1.5 of Cl

Now for empirical formula dividing all the moles by the smallest mole value and that is of Chlorine = 1.5

Carbon =    = 2

Hydrogen = = 4.2 = 4

Chlorine =    = 1

So empirical formula will be C₂H₄Cl

So option (A) C₂H₄Cl   is correct