I need help with some chemistry questions...?

Question

1. Consider the hydrocarbon pentane, C5H12 (molar mass 72.15 g).

(a) Write the balanced equation for the combustion of pentane to yield carbon dioxide and water.

(b) What volume of dry carbon dioxide, measured at 25°C and 785 mm Hg, will result from the complete combustion of 2.50 g of pentane?

(c) Under identical conditions, a sample of an unknown gas effuses into a vacuum at twice the rate that a sample of pentane gas effuses. Calculate the molar mass of the unknown gas.

2. Answer the following questions related to hydrocarbons.

(a) Determine the empirical formula of a hydrocarbon that contains 85.7 percent carbon by mass.

(b) The density of the hydrocarbon in part (a) is 2.0 g L/l at 509°C and 0.948 atm.

1. Calculate the molar mass of the hydrocarbon.

2. Determine the molecular formula of the hydrocarbon.

Answer 1

1. a) C5H12+8O2==>5CO2 +6 H2O
b) PV=nRT; plug in the numbers and i got 4.1L
c) square root of (M2/M1); and you get18.04g; M= molar mass

2. a) 85.7g/12.011g = 7.136mol C
14.3g/ 1.008g = 14.1865mol H
7.136/7.136= 1(C)
14.1865/ 7.136= 2(H)
CH2
b) 1.Molar Mass = dRT/P; 135.45g
2. 135.45/14.027= 9.656
10(CH2)= C10H20
C10H20

Answer 2

you seem to want an entire unit of chemistry in one question.

Here are some basics,

pentane combustion is C5H12 == CO2 + H2O

you can balance the equation

For the next part use stocihiometry to get moles of CO2 and then use the ideal gas equation to get thevolume of CO2

For the next part use Graham;s Law. The rae of effusion is inversely proportional to the sq. root of the molecular mass ratios.

For your second question, divide the relative mass by the molecular mass to get the molar ratio. Use the ideal gas equation to get the molar mass. Compare tot the EF to get the molecular formula