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4 years ago

Measurements show that unknown compound X has the following composition:

Chemistry

1 answer:

zvonat [6]4 years ago

6 0

Answer: The empirical formula of X is $C_{5}H_4O_2$ .

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 62.4 g

Mass of H = 4.19 g

Mass of O = 33.2 g

Step 1 : convert given masses into moles.

Moles of C= $\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{62.4g}{12g/mole}=5.2moles$

Moles of H= $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.19g}{1g/mole}=4.19moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{33.2g}{16g/mole}=2.1moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{5.2}{2.1}=2.5$

For H = $\frac{4.19}{2.1}=2$

For O = $\frac{2.1}{2.1}=1$

The ratio of C: H: O = 2.5 : 2 : 1

Converting it into simple whole number ratios by multiplying by 2:

Hence the empirical formula of X is $C_{5}H_4O_2$ .

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4 years ago

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Answer:

$\boxed{2.65}$

Explanation:

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$m = \text{2 tablets} \times \dfrac{\text{325 mg}}{\text{1 tablet}} = \text{750 mg}$

2. Moles of ASA

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$n = \text{750 mg} \times \dfrac{\text{1 mmol}}{\text{180.16 mg }} = \text{4.163 mmol}$

3. Concentration of ASA

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5. Solve for x

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6. Solve the quadratic equation.

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$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{Substituting values into the formula, we get}\\x = 0.002258\qquad x = -0.002591\\\text{We reject the negative value, so}\\x = 0.002258$

7. Calculate the pH

$\rm [H_{3}O^{+}]= x \, mol \cdot L^{-1} = 0.002258 \, mol \cdot L^{-1}\\\text{pH} = -\log{\rm[H_{3}O^{+}]} = -\log{0.002258} = \mathbf{2.65}\\\text{The pH of the solution is } \boxed{\textbf{2.65}}$

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