Solveeit Logo

Question

Question: How many formula units are contained in\(\text{0}\text{.67}\,\text{g}\,\text{CaO}\)?...

How many formula units are contained in0.67gCaO\text{0}\text{.67}\,\text{g}\,\text{CaO}?

Explanation

Solution

Hint Simplest form of an ionic compound is known as its formula unit.
Formula unit is the empirical formula of a compound that expresses the simplest whole number ratio of atoms of variable elements present in one molecule of the compound.
One mole is the amount of substance that contains 6.022×10236.022\times {{10}^{23}} atoms, and if the weight of a molecule is given the calculation of number of moles of a compound is takes through following formula –
noofmole weightmolecularweight\text{no}\,\text{of}\,\text{mole }\,\text{= }\dfrac{\text{weight}}{\text{molecular}\,\text{weight}}

Complete Step by step solution –
To calculate the formula unit in 0.67gCaO\text{0}\text{.67g}\,\text{CaO} in the first step we will calculate the number of moles of CaO\text{CaO} present in this given amount.
To calculate the molar mass of a compound we will add the atomic weight of all the elements.
molarmass =40 +16 =56 !!!! atomicweightofCa=40andO=16g\text{molar}\,\text{mass =}\,\text{40 +16 =}\,\text{56}\,\,\text{ }\\!\\!\\{\\!\\!\text{ atomic}\,\text{weight}\,\text{of}\,\text{Ca=40}\,\text{and}\,\text{O=16g}\\}
To calculate the number of moles, we will divide the given mass by molar mass

& \text{no}\,\text{of}\,\text{moles}\,\,\text{= }\dfrac{\text{weight}}{\text{atomic}\,\text{weight}} \\\ &\implies \text{}\dfrac{\text{0}\text{.67}}{\text{60}} \\\ &\implies \text{0}\text{.01195}\,\text{mole} \end{aligned}$$ In second step to determine the formula unit of$\text{CaO}$, we will multiply the number of moles of formula unit by $6.022\times {{10}^{\text{23}}}{}^{\text{formula}\,\text{unit}}/{}_{\text{mol}}$ $\begin{aligned} & \text{formula}\,\text{unit}\,\text{of}\,\text{CaO = }\,\text{0}\text{.01195}\,\text{mole }\\!\\!\times\\!\\!\text{ }\,\text{6}\text{.022 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}{}^{\text{formula}\,\text{unit}}/{}_{\text{mole}} \\\ & \text{formula}\,\text{unit}\,\text{of}\,\text{CaO}=\,7.2\times {{10}^{22}} \\\ \end{aligned}$ **Note –** Do not get confused with the molecular formula and empirical formula of the compound. The molecular formula of a compound represents the actual numbers of atoms present in one molecule of the compound, it implies that it shows the real formula of its one molecule. However empirical formula or formula unit is the simplest whole number ratio of atoms in the compound.