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Lyrx [107]
3 years ago
15

A compound is made up of 28 g N, 24 g C, 48 g O, and 8 g H .What is the empirical formula?

Chemistry
1 answer:
vovikov84 [41]3 years ago
6 0

Answer:

\rm C_2H_8N_2O_3.

Explanation:

<h3>Step One: calculate the coefficients. </h3>

Look up the relative atomic mass of these four elements on a modern periodic table:

  • \rm C: approximately 12.
  • \rm H: approximately 1.
  • \rm N: approximately 14.
  • \rm O: approximately 16.

The relative atomic mass of an element is numerically equal to the mass (in grams, \rm g,) of one mole of atoms of this element.

For example, the relative atomic mass of \rm C is approximately 12. Therefore, each mole of \rm C\! atoms would have a mass of 12\; \rm g.

This sample contains 24\; \rm g of carbon. That would correspond to approximately \displaystyle \left(\frac{24}{12}\right)\; \rm mol = 2\; \rm mol of \rm C atoms.

Similarly, for the other three elements:

\displaystyle n(\mathrm{H}) \approx \frac{8\; \rm g}{1\; \rm g \cdot mol^{-1}} = 8\; \rm mol.

\displaystyle n(\mathrm{N}) \approx \frac{28\; \rm g}{14\; \rm g \cdot mol^{-1}} = 2\; \rm mol.

\displaystyle n(\mathrm{O}) \approx \frac{48\; \rm g}{16\; \rm g \cdot mol^{-1}} = 3\; \rm mol.

Hence, the ratio between these elements in this compound would be:

n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3.

In the empirical formula of a compound, the coefficients should represent the smallest possible integer ratio between the number of atoms of these elements.

n(\mathrm{C}): n(\mathrm{H}): n(\mathrm{N}):n(\mathrm{O}) = 2: 8 : 2 : 3 is indeed the smallest possible integer ratio between the number of atoms of these elements.

<h3>Step Two: arrange the elements in an appropriate order</h3>

Apply the Hill System to arrange these four elements in the empirical formula. In the Hill System:

If carbon, \rm C, is present in this compound, then:

  • \rm C (carbon) and then \rm H (hydrogen) will be the first two elements listed in the formula (ignore the hydrogen if it is not in the compound.)
  • The other elements in this compound will be listed in alphabetical order.

If there is no carbon \rm C in this compound, then list all the elements in this compound in alphabetical order.

Both \rm C (carbon) and \rm H (hydrogen) are found in this compound. Therefore, the first element in the list would be \rm C\!. The second would be \rm H\!, followed by \rm N\! and then \rm O\!.

Hence, the empirical formula of this compound would be \rm C_2H_8N_2O_3.

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Identify the correct coefificients to balance it<br> -C3H8+O2 to -CO2 +-H2O
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Answer:

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Explanation:

{\rm ?\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Among the four species in this reaction, \rm C_3H_8 is species with the largest number of atoms per molecule. Assume that the coefficient of this compound is 1.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {?\; \rm CO_2} + {?\; \rm H_2O}.

Number of atoms on the left-hand side of the reaction:

  • \rm C: 1 \times 3 = 3.
  • \rm H: 1 \times 8 = 8.
  • \rm O: not found yet.

By the conservation of atoms, the number of atoms on the right-hand side of the reaction should match those on the left-hand side. In this reaction, \rm CO_2 is the only product with carbon atoms, whereas \rm H_2O is the only product with hydrogen atoms. These 3 carbon atoms and 8 hydrogen atoms would correspond to:

  • 3 / 1 = 3 \rm CO_2 molecules, and
  • 8 / 2 = 4 \rm H_2O molecules.

{\rm 1\; C_3H_8} + {\rm ?\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

Number of atoms on the right-hand side of the reaction:

  • \rm C: 3 \times 1 = 3.
  • \rm H: 4 \times 2 = 8.
  • \rm O: 3 \times 2 + 4 \times 1 = 10.

The number of \rm O atoms on the left-hand side should match those on the right-hand side. In this reaction, \rm O_2 is the only reactant with \rm O\! atoms. These 10 \rm \! O atoms would correspond to:

  • 5 \rm O_2 molecules.

{\rm 1\; C_3H_8} + {\rm 5\; O_2} \to {3\; \rm CO_2} + {4\; \rm H_2O}.

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A sample of a compound contains 160 g of oxygen and 20.2 g of hydrogen. Give the compound's empirical formula.
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Amount of oxygen in the compound = 160 g
Amount of oxygen in the compound = 20.2 gm
Mole of oxygen in the compound = 160/16
                                                     = 10 moles
Mole of hydrogen in the compound = 20.2/1.01
                                                         = 20 moles
Then
The ratio of oxygen to ration of hydrogen = 1:2
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