A sample of a gas contains 0.305 g of carbon, 0.407 g of oxygen and 1.805 g of chlorine. A different sample of the same gas having a mass of 8.84 g occupies 2 L at STP. What is the molecular formula for the compound

The molecular formula of the compound containing 0.305 g of carbon, 0.407 g of oxygen and 1.805 g of chlorine is COCl₂

We'll begin by calculating the empirical formula of the compound. This can be obtained as follow:

C = 0.305 g

O = 0.407 g

Cl = 1.805 g

Divide by their molar mass

C = 0.305 / 12 = 0.0254

O = 0.407 / 16 = 0.0254

Cl = 1.805 / 35.5 = 0.0508

Divide by the smallest

C = 0.0254 / 0.0254 = 1

O = 0.0254 / 0.0254 = 1

Cl = 0.0508 / 0.0254 = 2

Thus, the empirical formula of the compound is COCl₂

Next, we shall determine the number of mole of the gas that occupies 2 L at STP.

At standard temperature and pressure (STP),

22.4 L = 1 mole

Therefore,

2 L = 2 / 22.4

2 L = 0.089 mole

Thus, the number of mole of the gas that occupied 2 L at STP is 0.089 mole.

Next, we shall determine the molar mass of the gas.

Mass of gas = 8.84 g

Mole of gas = 0.089 mole.

Molar mass = mass / mole

Molar mass of gas = 8.84 / 0.089

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Empirical formula = COCl₂

Molar mass = 99.3 g/mol

Molecular formula = Empirical × n = molar mass

[COCl₂]n = 99.3

[12 + 16 + (2×35.5)]n = 99.3

99n = 99.3

Divide both side by 99

n = 99.3 / 99

Molecular formula = [COCl₂]n  

Molecular formula = [COCl₂] × 1

Therefore, the molecular formula of the compound is COCl₂

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Molecular formula = empirical formula  = COCl2

Explanation:

Step 1: Data given

MAss of carbon = 0.305 grams

MAss of oxygen = 0.407 grams

Mass of chlorine = 1.805 grams

A different sample of the same gas having a mass of 8.84 g occupies 2 L at STP

Atomic mass of C = 12.01 g/mol

Molar mass of O2 = 16.0 g/mol

Molar mass of Cl2 = 35.45 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 0.305 grams / 12.01 g/mol

Moles C = 0.0254 moles

Moles O = 0.407 grams / 16.0 g/mol

Moles O = 0.0254 moles

Moles Cl = 1.805 grams / 35.45 g/mol

Moles Cl = 0.0509 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 0.0254 moles / 0.0254 moles = 1

O: 0.0254 moles / 0.0254 moles = 1

Cl: 0.0509 / 0.0254 = 2

The empirical formula is COCl2

Step 4: Calculate moles

At STP we have 22.4 L for 1 mol

For 2 L we have 1 / 11.2 = 0.0893 moles

Step 5: Calculate the molar mass

Molar mass = mass / moles

Molar mass = 8.84 grams / 0.0893 moles

Molar mass = 98.99 g/mol

Step 6: Calculate the molar mass of the empirical formula

COCl2 = 12.01 + 16.0 + 70.9 ≠ 98.91 g/mol

Step 7: Calculate the molecular formula

Molecular formula = empirical formula  = COCl2

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.