An unknown organic compound composed of carbon, hydrogen and oxygen was analyzed and found to be 50.84% C, 8.53% H and the rest being oxygen. Which of the following represents the correct empirical formula for the compound? (a) CH2O
(b) C3H602
(c) C4H803
(d) C2H40
(e) C5H1003

e) C5H10O3

Explanation:

According to the information given in this question;

50.84% represents C

8.53% represents H

[100% - (50.84 + 8.53)]

= 100 - 59.37

= 40.63% of Oxygen (O)

This percentage means that;

Carbon = 50.84g

Hydrogen = 8.53g

Oxygen = 40.63g

By dividing by their respective atomic masses, we convert to moles:

C = 50.84g ÷ 12 = 4.24mol

H = 8.53g ÷ 1 = 8.53mol

O = 40.63g ÷ 16 = 2.54mol

Next we divide by the smallest no of the values (2.54mol)

C = 4.24 ÷ 2.54 = 1.669

H = 8.53 ÷ 2.54 = 3.358

O = 2.54 ÷ 2.54 = 1

To get a simple whole number ratio, we multiply the results by 3

C = 1.669 × 3 = 5.007

H = 3.358 × 3 = 10.074

O = 1 × 3 = 3

Simple whole number ratio of carbon, hydrogen and oxygen is 5:10:3. Hence, the empirical formula is C5H10O3.

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

a:

by the combination of an unlimited number of elements