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liq [111]
3 years ago
10

How many representative particles are in a 53.79 gram sample of SrCO3?

Chemistry
2 answers:
Tcecarenko [31]3 years ago
8 0
Remember representative particles/ molecules/ atoms are all the same 

<span>convert from grams -> moles </span>

<span>hope this clears thing</span>
Irina18 [472]3 years ago
5 0
To find the number of particles, you need to use Avogadro's number

6.02 x 10^23= 1 mol

As you can see, before we can convert to particles, we need moles. So let's first convert grams to moles using the molar mass (molecular weight) of the compound

molar mass of SrCO₃= 87.6 + 12.0 + (3 x 16.0)= 147.6 g/mol

53.79 g (1 mol/ 147.6 g)= 0.364 mol

now let's convert moles to particles using Avogadro's number

0.364 mol SrCO₃ (6.02x10^23/ 1 mol)= 2.19 x 10^23


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Step 3 - Finding the percent composition of carbon

As we saw in the previous step, the molar mass of Cr2(CO3)3 is 284 g/mol. From this molar mass, 36 g/mol come from C. We can set the following proportion:

\begin{gathered} 284\text{ g/mol ---- 100\%} \\ 36\text{ g/mol ----- x} \\  \\ x=\frac{36\times100}{284}=\frac{3600}{284}=12.7\text{ \%} \end{gathered}

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1 year ago
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