Question

How many moles are in \[550.0g\] of lead (II) sulfate?

Answer 1

Hint: Here we have to find the number of moles as we can convert from mole of a given substance to grams or vice versa by using molar mass concept of said substance. It means the actual concept we will use here is conversion of moles to gram or vice versa. Actually when we compare the amount of one particular substance to another using the mole concept , we should use grams also as such information can be provided from balances. To convert grams to mole , at first we have to determine how many grams are provided in the question (first step). Next , we have to calculate the molar mass of the substance(second step). The final step is to divide first step by second step.

Formula used: Molar mass \[ = \] mass in grams \[/\] \[1mole\] of substance

Complete step-by-step answer:
In this question we are actually using the concept which is concerned with conversion of moles to grams or vice versa by using molar mass . Here our aim is to find the number of moles in lead (II) sulfate or in general chemical formula state, which is \[PbS{O_4}\].

Before proceeding to calculation , let us have a look on to the concept of molar mass . We have to know that molar mass provides the mass of \[1mole\] of a particular given substance. The equation for molar mass is as follows,
Molar mass \[ = \] mass in grams \[/\] \[1mole\] of substance

We have to keep in mind that our case is lead (II) sulfate or \[PbS{O_4}\] which is having a molar mass of \[303.2626gmo{l^{ - 1}}\]. Actually from this we can understand that every mole of \[PbS{O_4}\] is having a mass of \[303.2626g\].

Next our duty is to convert from grams to moles so that enables us to set up molar mass as a conversion factor along with mass of \[1\,mole\] as a denominator which we find above. So, the conversion will be as follows,
\[\dfrac{{1\, mole \, PbS{O_4}}}{{303.2626g}}\]

Here, the given mass of \[PbS{O_4}\] is \[550.0g\]. Therefore,
\[\Rightarrow 550g \times \dfrac{{1\,mole\, PbS{O_4}}}{{303.2626g}}\]
\[\Rightarrow 1.8136\,mole\,PbS{O_4}\]
\[ \approx 1.814\,mole\,PbS{O_4}\]

Hence, there are \[1.814mole\] of \[PbS{O_4}\] is present in \[550.0g\] of lead (II) sulfate.

Note: \[PbS{O_4}\] is actually a white solid which appears as white color in microcrystalline form. This compound can be prepared by treating lead oxide , hydroxide or carbonate with hot sulfuric acid or otherwise by treating a soluble lead salt with sulfuric acid.

One main point we have to keep in mind is about the molar mass of \[PbS{O_4}\] because then only we can proceed in calculation as we did above. Next, we should remember to set up molar mass as a conversion factor with denominator of \[1mole\] mass while converting from grams to moles. Actually, same steps can be followed in these type questions regarding finding moles in grams of any given chemical compound.