Hello I understand How to do empirical problems, but the way this problem is worded has me so confused.
when 1.00g lithium metal is reacted with fluorine gas, the RESULTING COMPOUND has a mass of 3.73g. Calc the emprical fromula of lithium fluoride.
I know for lithium i would 1.00g/its molar mass
I am confused what would be fluorine would it be 99g? or 2.73? Please help me
1.00 g Li ... plus ... 2.73 g F
divide by their respective molar masses to find the molar ratio of the empirical formula
Remember the Law of Conservation of Mass. What you start with on the left must equal what you end uyp with on the right.
2Li + F2 ==> 2LiF
1.00. +x = 3.73 so what's x?
I can help you with that! To determine the empirical formula of lithium fluoride, we need to find the ratio of lithium to fluorine atoms in the compound.
First, let's calculate the moles of lithium using the given mass of 1.00g:
Molar mass of lithium (Li) = 6.94g/mol
Moles of lithium = mass of lithium / molar mass of lithium
= 1.00g / 6.94g/mol
≈ 0.144 mol
Next, let's determine the moles of fluorine by subtracting the mass of lithium from the resulting compound:
Mass of resulting compound = 3.73g
Mass of fluorine in the compound = Mass of resulting compound - Mass of lithium
= 3.73g - 1.00g
= 2.73g
Molar mass of fluorine (F) = 18.998g/mol
Moles of fluorine = mass of fluorine / molar mass of fluorine
= 2.73g / 18.998g/mol
≈ 0.144 mol
Now that we have the moles of both lithium and fluorine, we can determine their ratio. Since both moles are approximately equal to 0.144 mol, the ratio of lithium to fluorine in the compound is 1:1.
Therefore, the empirical formula of lithium fluoride is LiF, which means there is one lithium atom for every fluorine atom in the compound.
I hope this explanation helps you understand how to approach this empirical formula problem! Let me know if you have any further questions.