Weight of crucible , cover dish after heating = 67.092g Weight of magnesium , cr

Question

Weight of crucible , cover dish after heating = 67.092g

Weight of magnesium , crucible , cover and dish = 67.260 g

Weight of magnesium oxide , crucible , cover , dish after first heating = 67.340 g

Weight of magnesium oxide , crucible , cover , dish after second heating = 67.350 g

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1) what is the atomic mass of magnesium ?

2) what is the atomic mass of oxygen ?

3) weight of magnisium oxide ?

4) weight of magnesium in magnesium oxide ?

5)Moles of magnesium on magnesium oxide ?

6) Moles of oxygen in magnesium oxide ?

7)Element with smallar number of moles divided by itself ?

8) Element with the larger number of moles divided by element with the smallar number of moles ?

9) Empirical formula of magnesium oxide ?

10) Describe the appearance of the magnesium oxide ???

11) If you had used a different weight of magnesium metal , would the empirical formula of the compound be the same ? why or why not ???

Please show your work!

Thank You

Explanation / Answer

1. Atomicmass of Mg is 24.3 g/mol

2. Atomic mass of O is 16 g/mol

3. Weight of magnesium oxide = 67.350 - 67.092 = 0.258 g

4.

1 mole of MgO contains 1 mole Mg

0.258 g in moles = 0.258 g /40.3 g/mol = 6.40 x 10^-3 mol.

So, Mg =6.40 x 10^-3 mol.

Mg in grams = 6.40 x 10^-3 mol x 24.3 = 0.155 g

5.

Moles of magnesium in MgO = 6.40 x 10^-3 mol.

6.

1 mole of MgO has 1 mol of O.

So, moles of O in MgO = 6.40 x 10^-3 mol.

7.

Elements Mg and O are equal in moles.

So, moles of O divided by itself = 6.40 x 10^-3 mol./6.40 x 10^-3 mol. = 1

8.

Elements Mg and O are equal in moles.

So, moles of Mg divided by itself = 6.40 x 10^-3 mol./6.40 x 10^-3 mol. = 1

9.

Empirical formula = MgO

10.

It is a white color solid.

11.

The empirical formula would be same. Because, the mole ratios would be same, irrespective of weight of the element.

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