Calculate molecular weight composition

We know that 450mg of sodium are contained in a certain quantity of $Na_2S_2O_3$. How many grams of sulfur are there in the same quantity?

SOLUTION:
The molecular weight of $Na_2S_2O_3$ is:
$2\cdot23+2\cdot32.07+3\cdot16=158.14gr/mol$
The sodium percentage is:
$$\%Na=\frac{2\cdot23gr/mol}{158.14gr/mol}\cdot100=29.1\%$$

So, in 100gr of $Na_2S_2O_3$ there are 29.1gr of sodium; in $X$ grams of $Na_2S_2O_3$ there are 0.45gr of sodium:
$$100gr:29.1gr=X:0.45gr\Rightarrow X=\frac{0.45\cdot100}{29.1}=1.55gr$$

The sulfur percentage is:
$$\%S=\frac{2\cdot32.07gr/mol}{158.14}\cdot100=40.55\%$$

So, in 100gr of $Na_2S_2O_3$ there are 40.55gr of sulfur; in 1.55gr of $Na_2S_2O_3$ there are $X$ grams of sulfur:
$$100gr:40.55gr=1.55gr:X\Rightarrow X=\frac{40.55\cdot1.55}{100}=0.628gr=628mg$$

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PROBLEM 3:
3.5 grams of aluminium are contained in a certain quantity of aluminium sulfate. How many grams of oxygen are contained in the same quantity of the compound?
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Calculate molecular weight and the percent composition

Calculate the mass (in gr) of $Pb_3(PO_4)_3$ that contains $0.3gr$ of $Pb$. Then calculate the mass of $P$ contained in the same quantity of $Pb_3(PO_4)_3$. What is the percentage of oxygen?

SOLUTION:
Remember the atomic weight of the single atoms:
$\left\{\begin{matrix} Pb=207.2\\ P=31\\ O=16\\ \end{matrix}\right.$

The molecular weight of $Pb_3(PO_4)_3$ is:
$MW=207.2\cdot 3+\left ( 31+16\cdot 4 \right )\cdot 2=811.6gr/mol$
So one mole of $Pb_3(PO_4)_3$ has a mass of $811.6gr$.

The percentage of $Pb$ is:
$\%Pb=\frac{3\cdot 207.2gr/mol}{\displaystyle 811.6gr/mol}\cdot 100=76.59\%$
So, in $100gr$ of $Pb_3(PO_4)_3$ there are $76.59gr$ of $Pb$; $X$ grams of $Pb_3(PO_4)_3$ contains $0.3gr$ of $Pb$. Then:
$100gr:76.59gr=X:0.3gr\Rightarrow X=\frac{100gr\cdot 0.3gr}{76.59gr}=0.392gr=392mg$

The percentage of $P$ is:
$\%Pb=\frac{2\cdot 31gr/mol}{811.6gr/mol}\cdot 100=7.64\%$
So, in $100gr$ of $Pb_3(PO_4)_3$ there are $7.64gr$ of $P$; $0.392gr$ of $Pb_3(PO_4)_3$ contains $X$ grams of $P$. Then:
$100gr:7.64gr=0.392gr:X\Rightarrow X=\frac{7.64gr\cdot 0.392gr}{100gr}=0.0299gr=29.9mg$

The percentage of oxygen is:
$\%O=\frac{4\cdot 2\cdot 16gr/mol}{811.6gr/mol}\cdot 100=15.7\%$

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PROBLEM 2:
Calculate the mass of barium sulfate that contains 1.2gr of sulfur. Then compute the percentage of oxygen.
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How to calculate the percentage composition

Calculate the composition in percentage of the single atoms and $H_2O$ in the chemical compound $Na_2B_4O_7\cdot 10H_2O.$

SOLUTION:
First we consider the atomic weight of the species:
$\begin{cases} Na=23\\ B=10.8\\ O=16\\ H_2O=18\\ \end{cases}$

The molecular weight of $Na_2B_4O_7\cdot 10H_2O$ is:

$MW=2\cdot 23+4\cdot 10.8+7\cdot 16+10\cdot 18=381.2gr/mol$
So one mole of $Na_2B_4O_7\cdot 10H_2O$ has a mass of 381.2gr.

The percentage of the atoms in the compound is, in general:
$\%X=\frac{nX\cdot AtomicWeight}{AtomicMass}$

We can now calculate:
$\%Na=\frac{2\cdot 23gr/mol}{381.2gr/mol}\cdot 100=12.06\%$

$\%B=\frac{4\cdot 10.8gr/mol}{381.2gr/mol}\cdot 100=11.33\%$

$\%O=\frac{7\cdot 16gr/mol}{381.2gr/mol}\cdot 100=29.38\%$

$\%H_2O=\frac{10\cdot 18gr/mol}{381.2gr/mol}\cdot 100=47.22\%$

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PROBLEM 1:
Compute the percentage of K, Fe, C, N and $H_2O$ in $K_4Fe(CN)_6 \cdot 5H_2O$ crystals. Continue reading »