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cupoosta [38]
2 years ago
8

2-A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11%

Chemistry
1 answer:
aleksandrvk [35]2 years ago
3 0

The molecular formula :

C₆H₁₄O₃PF

<h3>Further explanation</h3>

Given

39.10% carbon, 7.67% hydrogen, 26.11%  oxygen, 16.82% phosphorus, and 10.30% fluorine.

Required

The molecular formula

Solution

mol ratio :

C = 39.1 : 12 = 3.258

H = 7.67 : 1 = 7.67

O = 26.11 : 16 = 1.632

P = 16.82 : 31 = 0.543

F = 10.3 : 19 = 0.542

Divide by 0.542

C = 6

H : 14

O = 3

P = 1

F = 1

The empirical formula :

C₆H₁₄O₃PF

(The empirical formula)n = the molecular formula

(C₆H₁₄O₃PF)=184.1

(6.12+14.1+3.16+31+19)n=184.1

(184)n=184.1

n = 1

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25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
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Answer:

Concentration of the original \rm KOH solution: approximately 1.05\; \rm mol \cdot dm^{-3}.

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\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}.

\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}.

Calculate the number of moles of \rm HCl molecules in that 0.0350\; \rm dm^{3} of 0.75\; \rm mol \cdot dm^{3} \rm HCl\! solution:

\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}.

\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

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