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How many atoms of each element is in \[CaC{O_3}\] ?

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Last updated date: 27th Feb 2024
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IVSAT 2024
Answer
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Hint: Molecule: The substance which is framed by similar kinds of components or elements for example same metals or same non-metal.
Compound: The substance which is framed by at least two distinct components or elements are known as compounds.

Complete answer:
Calcium carbonate is a synthetic inorganic compound having the substance equation formula \[CaC{O_3}\] .
Allow us first to discuss what molecules are.
Atoms: They are characterized as the littlest molecule of any substance. They can't be broken into pieces further. Molecules comprise three sorts of components for example electrons, protons and neutrons.
Element: It is characterized as the littlest piece of any substance. They are made of similar kinds of molecules. They are fundamentally of two sorts for example metals and non-metals. For instance: hydrogen, carbon, oxygen, and so on
Here we are given with the compound \[CaC{O_3}\] . Furthermore, we need to locate the all-out number of molecules present in the compound. From here we can see that there is a \[1\] calcium ( \[Ca\] ) atom present in this compound. Alongside calcium atoms carbon (represented by \[C\] ) and oxygen (represented by \[O\] ) are additionally present. Here the quantity of carbon atoms is \[1\] and the quantity of oxygen atoms in the compound is \[3\] . So, the absolute number of atoms present in the compound is \[1 + 1 + 3 = 5\] .
\[Ca{\text{ }} - {\text{ }}C{\text{ }} - {\text{ }}{O_3}\]
For the molar mass of calcium carbonate:
Add the nuclear mass of entire condition
\[Ca:{\text{ }}40\] , \[C:{\text{ }}12\] , \[O:{\text{ }}16\] since it is \[{O_3}\] times \[16\] by \[3\]
\[40 + 12 + 16\left( 3 \right) = 100{\text{ }}g/ml\]
\[100\] is the molar mass of the equation

Note: Atomic mass: Atomic mass is defined as the mass of the atom. It is represented by the symbol \[A\] .
Atomic number: It is characterized as the quantity of protons in an atom. It is represented by the symbol \[Z\] .
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