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1 year ago

What is the empirical formula of a substance that contains 0.117 mol of carbon, 0.233 mol of hydrogen, and 0.117 mol of oxygen?

1 answer:

6 0

Empirical formula of a compound gives the proportions of the elements in that compound but it does not define the actual arrangement and number of atoms.

Let the empirical formula of compound be $C_{x}H_{y}O_{z}$ .

The ratio of number of moles of C, H and O can be calculated as follows:

$C:H:O=0.117:0.233:0.117$

Simplifying the ratio,

$C:H:O=\frac{0.117}{0.117}:\frac{0.233}{0.117}:\frac{0.117}{0.117}=1:1.99:1=1:2:1$

Thus, the value of x, y and z will be 1, 2 and 1 respectively.

Therefore, the empirical formula will be $CH_{2}O$ .

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Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

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1 year ago

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When metals combine with nonmetals, the metallic atoms tend to 1. lose electrons and become positive ions 2. lose electrons and

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The correct answer is 1. Lose electrons and become positive ions.

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1 year ago

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The reaction H+ + HCO3â€"" <--> H2CO3 is a part of the respiration process. It takes place at _____ . A. the lungs and tis

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Answer:

The tissue cells

Explanation:

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$H^{+}+HCO^{3} ^{-} H_{2}CO_{3}$

It all starts from Carbondioxide. This Carbondioxide is dissolved in the blood and taken by red blood cell and converted into carbonic acid. It then dissociates to form a bicarbonate ion $HCO^{3} ^{-}$ and a hydrogen ion $H^{+}$

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2 years ago

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A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight

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Answer: The molecular formula will be $C_8H_8O_2$

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= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles$

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

For C = $\frac{5.9}{1.5}=4$

For H = $\frac{5.9}{1.5}=4$

For O = $\frac{1.5}{1.5}=1$

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is $C_4H_4O$

The empirical weight of $C_4H_4O$ = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2$

The molecular formula will be= $2\times C_4H_4O=C_8H_8O_2$

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2 years ago

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