Hi I got a question regarding the following calculations:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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

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

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

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 [itex]C_{3} H_{6}

Br_{2}[/itex]

Next I need to find the molecule's formula : [itex](C_{3} H_{6} Br_{2})_n[/itex]

This is done by calculating the n-value(which is the ratio between the moler masse of [itex](C_{3} H_{6} Br_{2})_n[/itex] 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.

[itex]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[/itex]

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

We can now calculate the n in [itex](C_{3} H_{6} Br_{2})_n:[/itex]

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

The the molecule's formula is then : [itex]C_{3} H_{6} Br_{2}[/itex]

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

Sincerley

Fred

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# An Organic substance

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