Answer to Question #138772 in General Chemistry for Sammy

Question

1. During the experiment, Althea prepares a specific chemical to test on. The chemical has 56% C, 18%H, 32% O. a. Predict the Empirical formula b. If the amount of that compound being used is 120 g/mole, what will be its molecular formula? *
2. Marco uses Ammonia (NH3) on his titration experiment. If the experimental usage of the NH3 is 68g/mole. What is the Molecular formula of of NH3 being used? *
3. Determine the empirical formulas for compounds with the following percent compositions: (a) 15.8% carbon and 84.2% sulfur (b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen *

Answer 1

1. (a) Moles of C = $=\dfrac{56}{12}=4.67$ ( mole ratio= 4.67/2 =2)

Moles of H$=\dfrac{18}{1} = 18$ ( mole ratio = 18/2 =9)

Moles of O $=\dfrac{32}{16}=2$ ( mole ratio = 2/2 = 1)

EF is therefore; C₂H₉O₁

₌ C₂H₉O

(b) Molecular formula = Empirical formula x n

$n=\dfrac{Molecular formular mass}{empirical formula mass} = \dfrac{120}{(24+9+16)}$ = 2.4 (approx. 2)

MF = (C₂H₉O₁) x 2

= C₄H₁₈O₂

2.

MF = EF x n

$n= \dfrac{68}{(14+3)} = 4$

MF =(NH₃) x 4

= N₄H₁₃

3.

(a) Moles of C $=\dfrac{15.8}{12} = 1.317$ ( Mole ratio = 1.317/1.317 = 1)

Moles of S $= \dfrac{84.2}{32.1} = 2.623$ (mole ratio = 2.623/1.317 = 2)

EF = C₁S₂

= CS₂

(b) Moles of C $=\dfrac{40}{12} = 3.333$ (mole ratio = 3.333/3.331 = 1)

Moles of H $=\dfrac{6.7}{1} = 6.7$ (mole ratio = 6.7/3.331 = 2)

Moles of O $=\dfrac{53.3}{16} = 3.331$ (Mole ratio = 3.331/3.331 = 1)

EF = C₁H₂O₁

= CH₂O