Combustion analysis of naphthalene, a hydrocarbon used in mothballs, produced 8.80 g co2 and 1.44 g h2o . calculate the empirical formula for naphthalene.

In a complete combustion process, the products would be 1 mol of CO2 and 1 mol of H2O. In order to find out the amount of naphthalene reacted in the combustion process, you would need the relation of the mass of carbon with carbon dioxide (1 mol of CO2 = 1 mol C) and also the mass of hydrogen with water (1 mol H2O = 2 mol H). Convert the mass of CO2 and H2O given to moles of Carbon and Hydrogen respectively. You would get 0.1998 mol C and 0.1598 mol H. Then you would need to divide each with the smallest amount of moles present which is 0.1598. Thus you would get a ratio of 1.25 C: 1 H. Multiply this ratio with 4 in order to make it whole number. The empirical formula would be C5H4. 

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