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A chemist finds that 30.82 g of nitrogen will react with 17.60 g, 35.20 g, 70.40 g, or 88.00 g of oxygen to form four different compounds. Calculate the mass of oxygen per gram of nitrogen in each compound.

A chemist finds that 3082 g of nitrogen will react with 1760 g 3520 g 7040 g or 8800 g of oxygen to form four different compounds Calculate the mass of oxygen p class=

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Answer : The mass of oxygen per gram of nitrogen for four different oxygen are, 0.57, 1.14, 2.28 and 2.85 grams respectively.

Explanation :

First we have to calculate the mass of oxygen per gram of nitrogen for oxygen 1.

As, 30.82 g of nitrogen will react with 17.60 g of oxygen

So, 1 g of nitrogen will react with [tex]\frac{17.60}{30.82}=0.57g[/tex] of oxygen

Now we have to calculate the mass of oxygen per gram of nitrogen for oxygen 2.

As, 30.82 g of nitrogen will react with 35.20 g of oxygen

So, 1 g of nitrogen will react with [tex]\frac{35.20}{30.82}=1.14g[/tex] of oxygen

First we have to calculate the mass of oxygen per gram of nitrogen for oxygen 3.

As, 30.82 g of nitrogen will react with 70.40 g of oxygen

So, 1 g of nitrogen will react with [tex]\frac{70.40}{30.82}=2.28g[/tex] of oxygen

First we have to calculate the mass of oxygen per gram of nitrogen for oxygen 4.

As, 30.82 g of nitrogen will react with 88.00 g of oxygen

So, 1 g of nitrogen will react with [tex]\frac{88.00}{30.82}=2.85g[/tex] of oxygen

Therefore, the mass of oxygen per gram of nitrogen for four different oxygen are, 0.57, 1.14, 2.28 and 2.85 grams respectively.

BladeRunner212
  • 0.57, 1.14, 2.28, and 2.85 grams respectively.
  • The latter compound does not comply with Dalton's law.

Further explanation

Let us calculate the mass of oxygen per gram of nitrogen in each compound.

First compound

[tex]\boxed{ \ \frac{17.60 \ g \ of \ O}{30.82 \ g \ of \ N} = 0.57\ }[/tex]

Second compound

[tex]\boxed{ \ \frac{35.20 \ g \ of \ O}{30.82 \ g \ of \ N} = 1.14\ }[/tex]

Third compound

[tex]\boxed{ \ \frac{70.40 \ g \ of \ O}{30.82 \ g \ of \ N} = 2.28\ }[/tex]

Fourth compound

[tex]\boxed{ \ \frac{88.00 \ g \ of \ O}{30.82 \ g \ of \ N} = 2.85\ }[/tex]

How do the numbers above support Dalton's atomic theory?

Let's find out the ratio that connects the results between the first, second, and third compounds above.

  • Between the first and second compounds: [tex]\boxed{ \ 0.57 \div 1.14 = \boxed{ \ 1 \div 2 \ } \ }[/tex]
  • Between the second and third compounds: [tex]\boxed{ \ 1.14 \div 2.28 = \boxed{ \ 1 \div 2 \ } \ }[/tex]
  • Between the first and third compounds: [tex]\boxed{ \ 0.57 \div 2.28 = \boxed{ \ 1 \div 2 \ } \ }[/tex]

It all obeys the law of multiple proportions because the ratio is simple whole numbers.

Notes:

  • The law of multiple proportions was put forward by John Dalton. This law is one of the basic laws of stoichiometry.
  • The law of multiple proportions states: if two elements can form more than one compound, the ratio of the mass of one element, which is compounded with a certain number of other elements, can be expressed in small whole-number ratios.

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Keywords: A chemist finds that 30.82 g of nitrogen, oxygen, different compounds, calculate the mass of oxygen per gram of nitrogen, Dalton, the law of multiple proportions.

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