# Question #7f838

##### 1 Answer

**!! LONG ANSWER !!**

Your compound's molecular formula is

This one is a little complex because there are many calculations to do before being able to determine the empirical and molecular formulas of the compound.

So, you know that your compound contains carbon, hydrogen, and oxygen. Moreover, you know that a **0.392-g** sample occupies a volume of **STP** - Standard Temperature and Pressure.

At STP, one mole of any ideal gas occupies exactly **22.4 L** - this is known as the molar volume of a gas at STP. This means that your sample contains

This implies that your compound's molar mass will be

Moving on. You know that this compound undergoes combustion and that **0.32 g** of water are produced. You'll use this value to determine how much *hydrogen* the initial sample contained.

Start by determining the percent composition of hydrogen in water

This means that, for every **100 g** of water, you get **11.19 g** of hydrogen. Therefore, for **0.32 g**, you get

Next, determine how much *carbon* your initial sample contained. Since the combustion reaction produces

Since you get **1 mole** of carbon for every **1 mole** of

Use carbon's molar mass to determine how many grams would contain this many moles

You can now figure out how much *oxygen* your initial sample had by

It's all downhill from here. Calculate how many moles of each element your sample contained by using their molar masses

Divide all these numbers by the smallest one to get the mole ratios of the elements

You compound's **empirical formula** will be

To get its molecular formula, you have to use the molar mass of the empirical formula and the molar mass of the compound.

Thus, your compound's **molecular formula** will be

**SIDE NOTE** *You get the same result by using the current definition of STP conditions, at which the volume occupied by an ideal gas is equal to 22.7 L, not 22.4 L.*

*However, I assume that your problem intended for you to use the old value, so I did the calculations using 22.4 L.*