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prohojiy [21]
3 years ago
14

How many atoms of hydrogen are present in 7.63 g of ammonia? please help me ??

Chemistry
1 answer:
GarryVolchara [31]3 years ago
3 0
The atoms  of  hydrogen  that are  present  in  7.63 g  of ammonia(NH3)

find  the  moles  of NH3 =mass/molar mass
 7.63 g/ 17 g/mol = 0.449  moles

since there  is  3 atoms of H  in NH3 the  moles of  H = 0.449 x 3 = 1.347 moles

by  use  of  1 mole = 6.02 x10^23  atoms
what  about  1.347  moles

= 1.347  moles/1   moles  x 6.02 x10^23 atoms = 8.11 x10^23  atoms of Hydrogen
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A 0.4647-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.01962 mol of
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Answer:

See explanation.

Explanation:

Hello,

In this case, we can show how the empirical formula is found by following the shown below procedure:

1. Compute the moles of carbon in carbon dioxide as the only source of carbon at the products:

n_C=0.01962molCO_2*\frac{1molC}{1molCO_2} =0.01962molC

2. Compute the moles of hydrogen in water as the only source of hydrogen at the products:

n_H=0.01961molH_2O*\frac{2molH}{1molH_2O}=0.03922molH

3. Compute the mass of oxygen by subtracting the mass of both carbon and hydrogen from the 0.4647-g sample:

m_O=0.4647g-0.01962molC*\frac{12gC}{1molC}-0.03922molH*\frac{1gH}{1molH}  =0.1900gO

4. Compute the moles of oxygen by using its molar mass:

n_O=0.1900gO*\frac{1molO}{16gO}=0.01188molO

5. Divide the moles of carbon, hydrogen and oxygen by the moles of oxygen (smallest one) to find the subscripts in the empirical formula:

C=\frac{0.01962}{0.01188}=1.65\\ \\H=\frac{0.03922}{0.01188} =3.3\\\\O=\frac{0.01188}{0.01188} =1

6. Search for the closest whole number (in this case multiply by 2):

C_3H_6O_2

Moreover, the empirical formula suggests this compound could be carboxylic acid since it has two oxygen atoms, nevertheless, this is not true since the molar mass is 222.27 g/mol, therefore, we should compute the molar mass of the empirical formula, that is:

M=12*3+1*6+16*2=74g/mol

Which is about three times in the molecular formula, for that reason, the actual formula is:

C_9H_{18}O_6

It suggest that the compound has a highly oxidizing character due to the presence of oxygen, therefore, we cannot predict the distribution of the functional groups as it could contain, carboxyl, carbonyl, hydroxyl or even peroxi.

Best regards.

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3 years ago
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