Respuesta :
Answer:
Mass percent of nitrogen in the compound is 13,3%
Explanation:
Dumas method is an analytical method to determine nitrogen content in samples, thus:
CₐHₓNₙ + (2a+x/2) CuO → aCO₂ + ˣ/₂ H₂O + ⁿ/₂ N₂ + (2a+x/2) Cu
As the CO₂ is removed with KOH, in the mixture you have H₂O and N₂
At 25°C. the vapor pressure of water is 23,8 torr, that means that the pressure due to N₂ is:
726torr - 23,8torr = 702,2 torr.
Using gas law:
n = PV/RT
Where:
P is pressure (702,2torr≡ 0,924 atm)
V is volume (0,0318L)
R is gas constant (0,082atmL/molK)
And T is temperature (25°C≡298,15K)
Replacing, number of moles of N₂, n, are:
n = 1,20x10⁻³moles of N₂. In grams:
1,20x10⁻³moles of N₂×[tex]\frac{28g}{1mol}[/tex] = 0,0336 g of N₂.
Thus, mass percent of nitrogen in the compound is:
[tex]\frac{0,0336g N}{0,253gSample}[/tex]×100= 13,3%
I hope it helps!
The mass percent of nitrogen in the compound is 13.3%
Dumas method:
- It is a quantitative determination of nitrogen in chemical substances.
The reaction involved will be:
CₐHₓNₙ + (2a+x/2) CuO → aCO₂ + ˣ/₂ H₂O + ⁿ/₂ N₂ + (2a+x/2) Cu
Since, CO₂ is removed with KOH, in the mixture thus only H₂O and N₂ are left.
- At 25°C, the vapor pressure of water is 23.8 torr.
Thus, the pressure due to N₂ is:
726torr - 23,8torr = 702.2 torr.
From ideal gas law:
[tex]n = PV/RT[/tex]
where:
- P is pressure (702.2torr≡ 0.924 atm)
- V is volume (0.0318L)
- R is gas constant (0,082atmL/molK)
- T is temperature (25°C≡298,15K)
n = 1.20x10⁻³moles of N₂. In grams:[tex]1.20*10^{-3} \text{ moles of } N_2*\frac{28g}{1mol} =0.0366g N_2[/tex]
Thus, mass percent of nitrogen in the compound is:
[tex]\frac{0.0366gN_2}{0.253g \text{ sample}} *100= 13.3\%[/tex]
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