# An Organic substance

1. Dec 2, 2004

### Mathman23

Hi I got a question regarding the following calculations:

An organic substance consisting of Carbon, Hydrogen and Bromine and has a total mass of 1,55 grams.

The carbon have a mass of 0,277 grams and the Hydrogen 0,046 grams.

The mass of Bromine is 1,55 g - 0,277 g - 0,046 g = 1,23 grams.

Next I calculate the number moles of Carbin, Hydrogen and Brome.

$n_{Carbon} = \frac{\textrm{0,277 g}}{\textrm{12,01 g/mole}} = \textrm{0,0230} \ \textrm{moles}$

$n_{Hydrogen} = \frac{\textrm{0,046 g}}{\textrm{1,008 g/mole}} = 0,046 \ \textrm{moles}$

$n_{Bromine} = \frac{\textrm{1,23 g}}{\textrm{79,9 g/mole}} = \textrm{0,0153} \ \textrm{moles}$

The mole ratio is a follows 0,023:0,046:0,0153 which can be reduced to 3:6:2.

The emperical formula for the substance is then $C_{3} H_{6} Br_{2}$

Next I need to find the molecule's formula : $(C_{3} H_{6} Br_{2})_n$
This is done by calculating the n-value(which is the ratio between the moler masse of $(C_{3} H_{6} Br_{2})_n$ and of the actual substance).

The actual molar mass is calculated as follows:
3,006 grams of the actual substance is placed in a container with a volume 500 mL. This is inturn heated to temperature of 150 degress celcius which generates a pressure within the container of 1,047 bars.

$n_{\textrm{actual substance}} = \frac{1,047 atm \cdot 0,500 L}{0,08206 \frac{L \cdot atm}{mol \cdot K}(150 + 273)K } = 0,0151 moles$

The molar masse for actual substance is then $M_{actual} = \frac{m_{actual}}{n_{actual}}= \frac{3,006 \ g}{0,0151 \ moles} = 199,073 g/mol$

We can now calculate the n in $(C_{3} H_{6} Br_{2})_n:$

$\frac{M_{actual}}{M_{(C_{3} H_{6} Br_{2})_n} = \frac{201,90 \cdot n }{199,08} : n = 1,014$

The the molecule's formula is then : $C_{3} H_{6} Br_{2}$

My question: Is this the correct approach to the above calculations ?

Sincerley
Fred

Last edited: Dec 2, 2004
2. Dec 2, 2004

I would use a fraction to represent each partial result (that you should use later in more complex calculations) and then approximate it. Take a look at the following:

$$n_{\mbox{Carbon}} = \frac{0.277 \mbox{ g }}{12.01 \mbox{ g/mole }} = \frac{277}{1210} \textrm{ mole } \approx 2.31 \times 10^{-2} \textrm{ moles }$$

Your approximation was 0.0230 moles (instead of 0.0231 moles). There are three mistakes:

1. Usage of comma vs. point (sometimes it's tricky):

- comma: 100,000 (one hundred thousand)
- point: 2.5 (two point five)
- comma & point: 100,000.56 (one hundred thousand and fifty-six)

NOTE: In South America, for example, people use the opposite notation. Points separate thousands, while a comma separate the integer part from the decimal one.

2. Let's go down one more decimal place. We approximately have: 0.02306. Since the last digit is $$N = 6 \geq 5$$, we must add 1 to our last decimal place. Otherwise, you would not need to add anything.

3. You may have noticed I used scientific notation up there. Try to stick to it.

Apply the same method to the remaining ones. Keep in mind it's better to use fractions when using results in further calculations.

Last edited: Dec 2, 2004