Nickel forms a compound with carbon monoxide $N{{i}_{x}}{{\left( CO \right)}_{y}}$. To determine its formula, you carefully heat a 0.0973g of sample in air to convert the nickel to 0.0426g of NiO and the CO to 0.100g of $C{{O}_{2}}$ . If the empirical formula is of the form $N{{i}_{x}}{{\left( CO \right)}_{y}}$[Ni = 58.7] then, x+y is:
Answer
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Hint: To solve this question firstly we have to calculate the molecular mass of the obtained products. After calculating the molecular mass, we have to compare it with the given weight of the products obtained and we will get a ratio of x and y from where we can find the required answer.
Complete step by step answer:
In the question, it is given to us that we heat 0.0973g of the sample of the compound that nickel forms with carbon monoxide. On heating, we get 0.0426g of NiO and 0.1g of carbon dioxide.
Also, the compound contains ‘x’ nickel atoms and ‘y’ carbon monoxide molecules. Therefore, following the stoichiometry, we can write the reaction as -
\[N{{i}_{x}}{{\left( CO \right)}_{y}}\xrightarrow{\Delta }xNiO+yC{{O}_{2}}\]
It is given to us that the atomic mass of nickel is 58.7.
Let us calculate the molecular mass of carbon monoxide. Atomic mass of carbon is 12 and atomic mass of oxygen is 16. Therefore, the molecular mass of carbon monoxide will be 12+ 16 i.e. 28.
Similarly for the products, molecular mass of NiO = atomic mass of nickel + atomic mass of oxygen = 58.7 + 16 = 74.7 and
Molecular mass of carbon dioxide = atomic mass of carbon + (2$\times $atomic mass of oxygen) = 12 + 32 = 44.
Now we can write the reaction as-
\[\begin{align}
& N{{i}_{x}}{{\left( CO \right)}_{y}}\xrightarrow{\Delta }xNiO+yC{{O}_{2}} \\
& (58.7x+28y)\to 74.7x+44y \\
\end{align}\]
The weight of NiO and carbon dioxide is given to us in the question which is 0.0426g and 0.1g respectively.
So we can write that –
$\begin{align}
& \dfrac{74.7x}{44y}=\dfrac{0.0426}{0.1} \\
& or,\dfrac{x}{y}=\dfrac{1.8744}{7.47}=0.25 \\
\end{align}$
I.e. the ratio of x and y is 0.25 which we can write in the fraction form as $\dfrac{1}{4}$ which means, $\dfrac{x}{y}=\dfrac{1}{4}$
Or, we can write that, x= 1 and y= 4.
So, x + y = 1+ 4= 5.
Therefore, the answer is x + y = 5.
Note: Here, since x is 1 and y is 4, the empirical formula of the compound will be $Ni{{(CO)}_{4}}$. Empirical formula of a compound shows us the simplest ratio between the atoms which are used in the formation of the atom.
Complete step by step answer:
In the question, it is given to us that we heat 0.0973g of the sample of the compound that nickel forms with carbon monoxide. On heating, we get 0.0426g of NiO and 0.1g of carbon dioxide.
Also, the compound contains ‘x’ nickel atoms and ‘y’ carbon monoxide molecules. Therefore, following the stoichiometry, we can write the reaction as -
\[N{{i}_{x}}{{\left( CO \right)}_{y}}\xrightarrow{\Delta }xNiO+yC{{O}_{2}}\]
It is given to us that the atomic mass of nickel is 58.7.
Let us calculate the molecular mass of carbon monoxide. Atomic mass of carbon is 12 and atomic mass of oxygen is 16. Therefore, the molecular mass of carbon monoxide will be 12+ 16 i.e. 28.
Similarly for the products, molecular mass of NiO = atomic mass of nickel + atomic mass of oxygen = 58.7 + 16 = 74.7 and
Molecular mass of carbon dioxide = atomic mass of carbon + (2$\times $atomic mass of oxygen) = 12 + 32 = 44.
Now we can write the reaction as-
\[\begin{align}
& N{{i}_{x}}{{\left( CO \right)}_{y}}\xrightarrow{\Delta }xNiO+yC{{O}_{2}} \\
& (58.7x+28y)\to 74.7x+44y \\
\end{align}\]
The weight of NiO and carbon dioxide is given to us in the question which is 0.0426g and 0.1g respectively.
So we can write that –
$\begin{align}
& \dfrac{74.7x}{44y}=\dfrac{0.0426}{0.1} \\
& or,\dfrac{x}{y}=\dfrac{1.8744}{7.47}=0.25 \\
\end{align}$
I.e. the ratio of x and y is 0.25 which we can write in the fraction form as $\dfrac{1}{4}$ which means, $\dfrac{x}{y}=\dfrac{1}{4}$
Or, we can write that, x= 1 and y= 4.
So, x + y = 1+ 4= 5.
Therefore, the answer is x + y = 5.
Note: Here, since x is 1 and y is 4, the empirical formula of the compound will be $Ni{{(CO)}_{4}}$. Empirical formula of a compound shows us the simplest ratio between the atoms which are used in the formation of the atom.
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