Organic Contains Carbon

History of carbon;the element that forms maximum no. of compounds

History of carbon                                             

  Carbon is one of the elements of great interest and importance. Its compounds are uncountable.  It forms the basis of life. Carbohydrates, proteins, fats etc. are all compounds of carbon.

Carbon shows catenation, a property by which it can form chains. This property allows  it to form the largest no. of compounds.

What is carbon basically?

 It is an element with atomic no. 6 and mass no. 12 for its normal isotope and 14 for its radioactive isotope. It has 6 electrons, 6protons and 6 or 8 neutrons.

Origin of carbon

 Burning organic compounds like kerosene yield carbon which is commonly called soot. It first originated by nuclear fusion in stars. All organic compounds and some inorganic compounds like carbonates, bicarbonates, carbon dioxide and carbon monoxide contain carbon in the combined state.

Allotropy

 It is the property by which an element occurs in different physical forms. The crystalline allotropes are diamond and graphite. Fullerene is a synthetic crystalline allotrope having atomicity 60.Among amorphous allotropes number lamp black or soot, gas carbon, coal etc. Coal is used as a fuel. Coke acts as a reducing agent in metallurgy. Water gas, producer gas contain CO and act as fuel. Diamond has a tetrahedral structure while graphite has a hexagonal layered structure having weak interplate bonds making it a good lubricant.

Chemical reactivity

 Graphite is thermodynamically the most stable allotrope. Diamond is unreactive in nature. Graphite is produced by thermal decomposition of SiC. Diamond dissolves in molten Fe. Coke is a very good reducing agent and is used to reduce heated metal oxides to metals. CO is also a reducing agent.

Some reactions involving carbon

 C+O2 →CO2

2C+O2→2CO

ZnO+C →Zn+CO

Fe2O3+3C →2Fe+3CO

CuCO3 →CuO+CO2

Uses of carbon

Diamond is the hardest substance. It scratches glass and is used for its cutting. Graphite is used as a lubricant. Lampblack is used to make shoe polish and printers ink. Coke is used as a reducing agent. Ethyne and oxygen produce high temperature flames. C14 is used to estimate the age of old objects by Radio Carbon Dating.Compounds like methane,butane are commonly used as fuels.

Find the empirical and molecular formula of an organic compound?

An organic compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compund yields 16.01 mg of Co2 and 4.37mg of H2O. The molar mass of the compound is 176.1 grams/mole. What are the empirical and molecular formulas of the compounds?

Well, let’s start by creating a chemical equation:

CxHxOx + O2 —-> CO2 + H2O

We have 16.01 mg of CO2. The molar mass of CO2 is 44.01g/mol. So we have 3.6 X 10^-4 moles of CO2.

We have 4.37mg of H2O. The molar mass of H2O is 18g/mol. So we have 2.4 X 10^-4 moles of H2O.

Since we’re only worried about the equation and we only care about the relationship between moles of CO2 and H2O, we can simplify this to

?CxHxOx + ?O2 —-> 3.6CO2 + 2.4H2O
Let’s deal with whole numbers before we go crazy here:

?CxHxOx + ?O2 —-> 36 CO2 + 24 H2O
Now simplify more, and since six is a multiple of both, we can get rid of it:

?CxHxOx + ?O2 —-> 6*6 CO2 + 6*4 H2O
?CxHxOx + ?O2 —-> 6 CO2 + 4 H2O

Looks like we can divide by two:
?CxHxOx + ?O2 —-> 3 CO2 + 2 H2O

That’s about as far as we’re going to get with this part.

What we know now is that our empirical molecular formula will need to have three carbons and four hydrogens, since we can pull O2 out of the air in a combustion reaction, but the carbon and hydrogen will have to come from our compound. So our formula needs to have a ratio of three carbons and for every four hydrogens. We don’t yet know how many oxygens we have in our molecular formula.

Our empirical formula will be C3H4O(unknown)

But we do have one more piece of information, the molar mass of the compound, 176.1g/mol. Here are the building blocks that will add up to that number, our atomic masses: C 12.01g/mol, O 16.0 g/mol, H 1.01 g/mol.

The ratio of carbon and hydrogen will need to stay three to four.

12.01 times three =36.03g/mol C
1*4=4g/mol H
Total mass of C and H is 40.03g

The rest is oxygen, so let’s see if that can fit: 176-40.03=135.97. Divide by oxygen’s molar mass of 16 to get 8.5. Can’t have a half a molecule. Let’s double everything.

12.01 times six=72.06g/mol C
1*8=8g/mol H
Total mass of C and H is 80.06g
176-80.06 = 95.94

95.94/16 is 5.9999 – I’ll take it. Sold. We have 6 oxygens. Our molecular formula is C6H8O6.

WHEW!

So to get the empircal formula, just simplify this by taking out any common multipliers to get C3H4O3.

Your answers are these:
The empirical formula is C3H4O3.
The molecular formula is C6H8O6.

Radiocarbon Dating

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