Magnesium oxide smoke lost during the reaction?

[Cuisine123]

Homework Statement

Would the Mg : O ratio be affected if some magnesium oxide smoke escaped during the reaction of Mg with oxygen gas?

Please explain with a sample calculation. Thanks.

Homework Equations

I know that the empirical formula is MgO.

The Attempt at a Solution

I don't understand how to do this question., but I think that the ratio will remain unchanged?

 

[chemisttree]

I assume you are calculating the ratio by weighing the magnesium ribbon before combustion and then weighing the ash after combustion. If some ash mass is lost, what will this do to your calculation?

You need to set up a thought experiment and then remove a little of the mass of ash and recalculate. What happens to the ratio?

 

[Cuisine123]

I assume you are calculating the ratio by weighing the magnesium ribbon before combustion and then weighing the ash after combustion. If some ash mass is lost, what will this do to your calculation?

You need to set up a thought experiment and then remove a little of the mass of ash and recalculate. What happens to the ratio?

I have already done an experiment to find out the empirical formula for Magnesium Oxide, but this is a follow-up question..If some ash mass is lost, then I guess the numbers to calculate the ratio would be smaller?
But since MgO is already reduced, then I guess the ratio will remain unchanged?

Also, how can I show it with a sample calculation? Thanks.

 

[chemisttree]

When you weighed the MgO, the balance didn't know it was MgO. All the balance tells you is how much mass is in the pan. YOU assign meaning to the reading.

 

[Cuisine123]

When you weighed the MgO, the balance didn't know it was MgO. All the balance tells you is how much mass is in the pan. YOU assign meaning to the reading.

Sorry, I still don't understand.

 

[Borek]

When doing calculations for th eesxperiment, you are making two assumptions - even if you are not aware of the fact that you are doing them.

First, you assume that the product that you have weighted is pure MgO.

Second, you assume that all Mg was converted to the oxide and none was lost.

Do you see now?

As for the calculations:

If there were some MgO lost in the form of "smoke", final mass of the product will be smaller. Subtract 10% from the mass you have found and repeat the calculations identical to those you have already performed. What ratio did you get this time? Is it identical to the one you got previous time?