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2 years ago

0.475 g H, 7.557 gS, 15.107 g O. Express your answer as a chemical formula.

Chemistry

1 answer:

max2010maxim [7]2 years ago

7 0

Answer:

H₂SO₄

Explanation:

We have a compound formed by 0.475 g H, 7.557 g S, 15.107 g O. In order to determine the empirical formula, we have to follow a series of steps.

Step 1: Calculate the total mass of the compound

Total mass = mass H + mass S + mass O = 0.475 g + 7.557 g + 15.107 g

Total mass = 23.139 g

Step 2: Determine the percent composition.

H: (0.475g/23.139g) × 100% = 2.05%

S: (7.557g/23.139g) × 100% = 32.66%

O: (15.107g/23.139g) × 100% = 65.29%

Step 3: Divide each percentage by the atomic mass of the element

H: 2.05/1.01 = 2.03

S: 32.66/32.07 = 1.018

O: 65.29/16.00 = 4.081

Step 4: Divide all the numbers by the smallest one

H: 2.03/1.018 ≈ 2

S: 1.018/1.018 = 1

O: 4.081/1.018 ≈ 4

The empirical formula of the compound is H₂SO₄.

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a. ppb of trichloroethylene = 3 × 10⁶ ppb

b. ppm of Cl₂ = 3.8 ppm

c. Molarity = 0.0002 mol / L

d. Molarity = 0.0007 mol / L

e. For trace amount of concentrations

Explanation:

a. Given data

mass of trichloroethylene = 25 mg

Volume of water = 9.5 L

ppb of trichloroethylene = ?

Solution

As we know that

1 L = 1000 milliliters

9.5 L = 9.5 × 1000

9.5 L = 9500 millileters (ml)

we consider 25 mg = 25 millileters

ppb = (mass of solute / mass of solvent) × 1000,000,000 (1 billion)

ppb of trichloroethylene = (25 ÷ 9500) × 1000,000,000

ppb of trichloroethylene = 0.003 × 1000,000,000

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B. Given data

Mass of Cl₂ = 38 g

volume of water = 1.00 × 10⁴ L ( 10000 L)

ppm of Cl₂ = ?

Solution

Volume of water in ml = 1 L = 1000 ml

Volume of water in ml = 10000 × 1000

Volume of water in ml = 10000000 ml

we take 38 g = 38 ml

Now we convert it to ppm

ppm = (mass of solute / mass of solvent) × 1000000 (1 million)

ppm of Cl₂ = ( 38 ÷ 10000000 ) × 1000000

ppm of Cl₂ = 0.0000038 × 1000000

ppm of Cl₂ = 3.8 ppm

C. Given data

Concentration of F⁻ ( Fluoride ion) = 2.4 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

2.4 ppm = 2.4 × 0.001 g/L

2.4 ppm = 0.0024 g/L

Mass of flouride ions = 0.0024 g

Now we find number of moles

moles = mass / molar mass

molar mass of F⁻ = 19 g/mol

moles of F⁻ = 0.0024 g / 19 g/mol

moles of F⁻ = 0.0002 mol

Molarity = mol of solute / liter of solution

Molarity = 0.0002 mol / 1 L

Molarity = 0.0002 mol / L

D. Given data

Concentration of NO₃⁻ ( nitrate ion) = 45 ppm

Molarity = ?

Solution

As we know that 1 ppm = 0.001 g / L

45 ppm = 45 × 0.001 g/L

45 ppm = 0.045 g/L

Mass of nitrate ions = 0.045 g

Now we find number of moles

moles = mass / molar mass

molar mass of NO₃⁻ = 62 g/mol

moles of NO₃⁻ = 0.045 g / 62 g/mol

moles of F⁻ = 0.0007 mol

Molarity = mol of solute / liter of solution

Molarity = 0.0007 mol / 1 L

Molarity = 0.0007 mol / L

E. Reason of expressing concentration in ppm and ppb

Scientist prefer ppm and ppb notations when the concentration difference of solute and solvent are very high.

As water contains contaminants is a very low amount we can say in trace amounts so scientist prefer ppm and ppb rather than molarity.

Example

Arcenic is an under ground water contaminant and its concentration of 10 μg/L is dangerous for health.

Lets change this in to molarity

mass = 10 μg

10 μg = 10 / 1000000

10 μg = 0.00001 g

now find out moles of Arcenic

moles = mass / molar mass

molar mass of arcenic = 75 g/mol

moles = mass / molar mass

moles of arcenic = 0.00001 g / 75 g/mol

moles of arcenic = 0.00000012 mol

Molarity = moles of solute / litres of solution

Molarity = 0.00000012 mol / 1 L

Molarity = 0.00000012 mol/ L

As we can see that in molarity it is a negligible amount so scientists express it in ppm and ppb

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