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2Na(s)+Cl 2 (g) 2NaCl(s)

A chemist determines that the mass of an empty flask with a stopper is 156.1870g. After filling the stoppered flask with chlorine gas (Cl2 ) it’s mass is 158.1743. The chemist introduces 0.3870g of elemental sodium (Na) into the flask, causing a violent reaction and creating NaCl. All of the solid sodium is used up by the reaction. The chemist collects the unreacted chlorine gas and determines its mass to be 0.7962 g.

1. How many moles of elemental sodium were used in the reaction?

2. What is the mass of chlorine gas used in the reaction?

3. How many moles of chlorine were used in the reaction?

4. What is the empirical formula of sodium chloride based on the experimental data?

5. Was the empirical formula you obtained correct using the chemists data correct? Why?

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Answer from: eiza666

1. 0.1683 mol2. 1.191 g3. 0.02695 mol4. Na₂Cl₃5. The empirical formula obtained is not correct. This is likely due to experimental errors, since much precision was required (the masses are determined in thousandths of grams).

Explanation:

1. How many moles of elemental sodium were used in the reaction?

Since all of the solid sodium is used up by the reaction, you can cancluate the number of moles of elemental sodium used dividing the mass by the molar mass:

number of moles = mass in grams / atomic massmass in grams = 0.3870 g (given)atomic mass = 22.990 g/molnumber of moles = 0.3870 g / 22.990 g/mol = 0.1683 mol

2. What is the mass of chlorine gas used in the reaction?

a) Mass of chlorine gas introduced in the flask = mass of the stoppered flask after filling it with chlorine gas - mass of the empty flask with the sopper

Mass of chlorine gas introduced = 158.1743g - 156.1870g = 1.9873 g

b) Mass of chlorine gas unreacted = 0.7962 g (given)

c) Mass of chlorine gas used = mass of chorine gas introduced in the flask - mass of chlorine gass un reacted

Mass of chlorine gas used = 1.9873g - 0.7962g = 1.1911g

3. How many moles of chlorine were used in the reaction?

molar mass of chlorine gas, Cl₂ = 2 × 35.453 g/mol = 70.906 g/molnumber of moles = mass in grams / molar mass = 1.911g / 70.906g/mol = 0.02695 mol

4. What is the empirical formula of sodium chloride based on the experimental data?

Divide the number of moles of each element by the smalles number of moles:

Na: 0.01683 / 0.01683 = 1Cl = 0.02695 / 0.01683 = 1.6

Multiply by 2 to obtain whole numbers:

Na = 2Cl = 3.2 = 3Empirical formula Na₂Cl₃

5. Was the empirical formula you obtained correct using the chemists data correct? Why?

No, the empirical formula you obtained using the chemists data is not correct, because the correct empirical formula of sodium chloride is NaCl.

That is, there is 1 atom of sodium per every atom of chlorine in one chemical formula of NaCl, but that is not reflected by the empirical formula Na₂Cl₃.

That is a demostration of big experimental errors. You can speculate that the errors are likely due to problems of procedure collecting the gas or errors in measuring the masses.

As you see, the masses are measured to thousandths of grams, which requires much precision; thus smalls absolute errors could produce huge relative errors.

Answer from: Quest

a solution is always transparent, light passes through with no scattering from solute particles which are molecule in size. the solution is homogeneous and does not settle out. a colloid is intermediate between asolution and a suspension. while asuspension will separate out a colloid will not.

What’s is a suspension, and how is it different from a solution?

Answer from: Quest

What’s the solution?

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2Na(s)+Cl 2 (g) 2NaCl(s)

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