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nataly862011 [7]
3 years ago
12

5.00 g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 7

8./gmol, is burned completely in excess oxygen, and the mass of the products carefully measured, product mass:
carbon dioxide 16.93g

water 3.46g

Use this information to find the molecular formula of X.
Chemistry
1 answer:
slega [8]3 years ago
4 0

Answer:

The molecular formula = C6H6

Explanation:

Step 1: Data given

Mass of compound X = 5.00 grams

Mass of products =

  CO2 = 16.39 grams

  H2O = 3.46 grams

Molar mass CO2 = 44.01 g/mol

Molar mass H2O = 18.02 g/mol

Molar mass C = 12.01 g/mol

Molar mass H = 1.01 g/mol

Molar mass O = 16. 0g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles CO2 = 16.93 grams / 44.01 g/mol

Moles CO2 =  0.385 moles

Moles C = 1* 0.385 = 0.385 moles

Moles H2O = 3.46 grams / 18.02

Moles H2O = 0.192 moles

Moles H = 2* 0.192 = 0.384 moles

Step 3: Calculate mass

Mass = moles * molar mass

Mass C = 0.385 moles *12.0 g/mol

Mass C = 4.62 grams

Mass H = 0.39 grams

Mass O = 5.00 - 4.62 -0.38 moles

Mass O = 0 grams

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.385 moles / 0.384 = 1

H: 0.384 moles / 0.384 = 1

The empirical formula is CH

This molecular formula is 13 g/mol

We have to multiply the empirical formula by n

n = 78 g/mol / 13 g/mol

n = 6

The molecular formula = 6*(CH) = C6H6

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2 years ago
Suppose a group of volunteers is planning to build a park near a local lake. The lake is known to contain low levels of arsenic
ale4655 [162]

A) 10.75 is the concentration of arsenic in the sample in parts per billion .

B) 7,633.66 kg the total mass of arsenic in the lake that the company have to remove.

C)It will take 1.37 years to remove all of the arsenic from the lake.

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A) Mass of arsenic in lake water sample = 164.5 ng

The ppb is the amount of solute (in micrograms) present in kilogram of a solvent. It is also known as parts-per million.

To calculate the ppm of oxygen in sea water, we use the equation:

\text{ppb}=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^9

Both the masses are in grams.

We are given:

Mass of arsenic = 164.5 ng = 164.5\times 10^{-9} g

1 ng=10^{-9} g

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Density of the lake water sample ,d= 1.00 g/cm^3

Mass of sample =  M = d\times V=1.0 g/cm^3\times 15.3 cm^3=15.3 g

ppb=\frac{164.5\times 10^{-9} g}{15.3 g}\times 10^9=10.75

10.75 is the concentration of arsenic in the sample in parts per billion.

B)

Mass of arsenic in 1 cm^3  of lake water = \frac{164.5\times 10^{-9} g}{15.3}=1.075\times 10^{-8} g

Mass of arsenic in 0.710 km^3 lake water be m.

1 km^3=10^{15} cm^3

Mass of arsenic in 0.710\times 10^{15} cm^3 lake water :

m=0.710\times 10^{15}\times 1.075\times 10^{-8} g=7,633,660.130 g

1 g = 0.001 kg

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C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

C)

Company claims that it takes 2.74 days to remove 41.90 kilogram of arsenic from lake water.

Days required to remove 1 kilogram of arsenic from the lake water :

\frac{2.74}{41.90} days

Then days required to remove 7,633.66 kg of arsenic from the lake water :

=7,633.66\times \frac{2.74}{41.90} days=499.19 days

1 year = 365 days

499.19 days = \frac{499.19}{365} years = 1.367 years\approx 1.37 years

It will take 1.37 years to remove all of the arsenic from the lake.

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