Date Lab Sec etg Part B Part C Part D 41.543 24.00 95 49.575 Mass of crucible an

Question

Date Lab Sec etg Part B Part C Part D 41.543 24.00 95 49.575 Mass of crucible and lid (g) 2. Mass ofcrucible, lid and sample (g) 3. Mass of sample (g) 4. Mass of crucible, lid and product 0.48 41.57425 25.737a gala 0.03 10429501 1st mass measurement (g) 2nd mass measurement (g) 3rd mass measurement (g) 115 7Lea 25.83 50.537 4,5775,(o 73 50.5a 8 41578 25 (o73g 50.5a 25.0780 50.533g 0.958 5. Final mass ofcrucible, lid, andproduct (g)41.577 6. Mass of product (g) 7. Part B. Combination Reaction of Magnesium and Oxygen a. Mass ratio of Mg and O b. Mole ratio of Mg and O c. Consensus empirical formula of magnesium oxide d. Percent by mass (%) %Mg;

Explanation / Answer

The first thing that you should notice here is that the problem provides you with the mass of magnesium and the mass of the oxide.

Since you know that

magnesium + oxygen gas magnesium oxide

you can say that the mass of oxygen gas that took place in the reaction must be equal to the difference between the mass of oxide and the mass of magnesium.

a.This means that you have

mass of oxide(4.03 g) mass of magnesium(2.3 g) = mass of oxygen gas(1.7 g)

Since you subtracted two values to get this answer, the mass of oxygen gas must be rounded to one decimal place because that's how many decimal places you have for your least precise measurement, i.e. for 2.3 g.

Mg/O = 2.3g/1.7g

b. Now, to find the number of moles of oxygen present in the oxide, you need to use the molar mass of elemental oxygen, O, not of diatomic oxygen, O2.

This is the case because the oxide does not contain any oxygen molecules, it only contains oxygen atoms!

1.7g1 mole O16.0g=0.10625 moles O

Do the same for magnesium--use its molar mass to find the number of moles of magnesium present in the oxide.

2.3g1 mole Mg24.305g=0.09463 moles Mg

To find the mole ratio of the two elements in the oxide, simply divide the two values by the smallest one.

For Mg: 0.09463moles0.09463moles=1

For O: 0.10625moles0.09463moles= 1.12

Mg/O = 1/1.12

c.

Now, in order to find the empirical formula of the oxide, you should use the smallest whole number ratio that exists between the two elements in the oxide.

In this case, you can't really come up with a whole number ratio unless you multiply both values by 9, in which case you'd end up with

Mg(19)O(1.129)Mg9O10

Ideally, the two elements should combine in a 1:1 mole ratio in the oxide, so you can say that a number of experimental/procedural errors have crept into your experiment.

d. Mass Percentage is

Mg/O = 0.023/0.017

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