How many grams of MgC03 contain 24-00 g of oxygen? (The molar mass of MgC03 is 84-30 g mol-1)

1. Basics of Matter: Elements and Compounds (basic)

At the very heart of the physical world around us is matter—anything that has mass and occupies space. To understand matter, we must look at its simplest building blocks: atoms. As you explore the nature of substances, you'll find that all matter is classified based on how these atoms are organized. An element is a pure substance consisting of only one type of atom. For instance, a block of pure gold contains nothing but gold atoms. Interestingly, many elements like Oxygen (O₂), Nitrogen (N₂), and Sulphur (S₈) do not exist as lone atoms in nature; instead, they bond with their own kind to form molecules Science, Class VIII, Particulate Nature of Matter, p.115. Whether it is a single atom or a diatomic molecule like Nitrogen, which forms a strong triple bond between its atoms, it remains an element because it is chemically uniform Science, Class X, Carbon and its Compounds, p.60. When atoms of different elements join together in a fixed, definite ratio, they form a compound. A classic example is water (H₂O), where two hydrogen atoms are chemically bonded to one oxygen atom. This is not just a simple mix; it is a chemical transformation. Once a compound is formed, the original elements lose their individual properties. For example, while oxygen helps things burn, the compound water is used to extinguish fires! Because these atoms are so tightly attached, you cannot separate a compound into its constituent elements by physical means like filtration or evaporation Science, Class VIII, Nature of Matter: Elements, Compounds, and Mixtures, p.124.

To help you distinguish between the two, consider this comparison:

Feature	Element	Compound
Composition	Only one type of atom.	Two or more different types of atoms.
Separation	Cannot be broken down further by chemical or physical means.	Can be broken down into elements only by chemical reactions.
Properties	Reflects the property of its constituent atom.	Properties are entirely different from the constituent elements.

Sources: Science, Class VIII (NCERT 2025), Particulate Nature of Matter, p.115; Science, Class X (NCERT 2025), Carbon and its Compounds, p.60-61; Science, Class VIII (NCERT 2025), Nature of Matter: Elements, Compounds, and Mixtures, p.124

2. Atomic Mass and Molecular Mass (basic)

To understand chemistry at its core, we must first understand how we weigh the building blocks of matter. Since individual atoms are too small to weigh on a standard scale, scientists use the Atomic Mass, which is a relative measure. For example, the atomic mass of Carbon is 12 u (unified atomic mass units) and Hydrogen is 1 u Science, class X (NCERT 2025 ed.), Carbon and its Compounds, p.66. These values represent the average mass of the atoms of an element compared to 1/12th the mass of a carbon-12 atom. Moving a step higher, the Molecular Mass of a substance is the sum of the atomic masses of all the atoms present in a single molecule of that substance. To calculate it, you simply multiply the atomic mass of each element by the number of atoms of that element present in the chemical formula and add them all together. For instance, in a molecule like Water (H₂O), the molecular mass is calculated as (2 × 1 u) + (1 × 16 u) = 18 u. This principle applies even to complex series of compounds, such as alcohols, where each successive member differs by a specific unit like –CH₂–, leading to a predictable increase in molecular mass Science, class X (NCERT 2025 ed.), Carbon and its Compounds, p.67. Understanding molecular mass is vital because it tells us the proportions of different elements within a compound. If a compound contains multiple atoms of a single element—such as the four oxygen atoms in Fe₃O₄—we must account for all of them when balancing reactions or calculating total mass Science, class X (NCERT 2025 ed.), Chemical Reactions and Equations, p.4. This ratio is constant for any pure sample of that compound, allowing us to determine exactly how much of an element is contained within a specific mass of a substance.

Sources: Science, class X (NCERT 2025 ed.), Carbon and its Compounds, p.66; Science, class X (NCERT 2025 ed.), Carbon and its Compounds, p.67; Science, class X (NCERT 2025 ed.), Chemical Reactions and Equations, p.4

3. The Mole Concept and Molar Mass (intermediate)

In the world of chemistry, we deal with entities so small that counting them individually is impossible. To bridge the gap between the microscopic world of atoms and the macroscopic world of the laboratory, we use the Mole Concept. Just as the word 'dozen' represents 12 items, one Mole represents exactly 6.022 × 10²³ particles (atoms, molecules, or ions). This specific number is known as Avogadro’s Number. The beauty of the mole is that it allows us to weigh out a substance in grams and know exactly how many atoms we have held in our hands.

The link between the two worlds is Molar Mass—the mass of one mole of a substance, expressed in grams per mole (g/mol). Numerically, the molar mass of an element is equal to its atomic mass. For example, as we see in Science, Class X (NCERT 2025 ed.), Carbon and its Compounds, p.66, the atomic mass of Carbon is 12 u and Hydrogen is 1 u. Therefore, the molar mass of Carbon is 12 g/mol. This means if you weigh out 12 grams of Carbon, you have exactly one mole of Carbon atoms.

When dealing with compounds, we calculate the Molecular Mass by summing the atomic masses of all the atoms in the chemical formula. For instance, consider a molecule like Methane (CH₄). It consists of one Carbon atom (12 u) and four Hydrogen atoms (4 × 1 u), giving it a molecular mass of 16 u and a molar mass of 16 g/mol. We can apply this logic to see how molecules change within a series; for example, every successive member of the alkane family differs by a –CH₂– unit, which corresponds to an increase in molecular mass of 14 u (12 for C + 2 for H), a concept explored when comparing homologous series in Science, Class X (NCERT 2025 ed.), Carbon and its Compounds, p.67.

Term	Definition	Units
Atomic Mass	The mass of a single atom.	u (unified atomic mass unit)
Molar Mass	The mass of 6.022 × 10²³ particles of a substance.	g/mol (grams per mole)

Sources: Science, Class X (NCERT 2025 ed.), Carbon and its Compounds, p.66; Science, Class X (NCERT 2025 ed.), Carbon and its Compounds, p.67

4. Law of Constant (Definite) Proportions (intermediate)

The Law of Constant Proportions (also known as the Law of Definite Proportions) is a pillar of chemistry first proposed by Joseph Proust. It states that a chemical compound, no matter its source or how it was prepared, always contains the same elements combined together in the same proportion by mass. This means a compound isn't just a random mix; it has a fixed chemical "identity." For example, pure water (H₂O) will always consist of hydrogen and oxygen in a mass ratio of approximately 1:8, whether it comes from a tap in Delhi or a glacier in the Antarctic.

This principle distinguishes compounds from mixtures. While you can mix sand and salt in any ratio you like, elements in a compound are chemically bonded in specific, unchanging ratios. As noted in Science, Class VIII NCERT, Nature of Matter: Elements, Compounds, and Mixtures, p.124, substances like sodium chloride (common salt) are made of particles in a fixed 1:1 atomic ratio. Because each atom has a specific mass, this fixed atomic ratio inevitably leads to a fixed mass ratio for the entire compound.

Understanding this law is crucial for stoichiometry—the calculation of quantities in chemical reactions. If we know the percentage of one element in a compound, we can predict exactly how much of that compound we need to obtain a specific amount of that element. For instance, if you are analyzing magnesium carbonate (MgCO₃), the law ensures that the ratio of magnesium to carbon to oxygen is eternally fixed. If you double the amount of the compound, you double the mass of each constituent element, but the percentage of each remains exactly the same.

Sources: Science, Class VIII NCERT, Nature of Matter: Elements, Compounds, and Mixtures, p.124

5. Chemical Formulas: Interpreting MgCO₃ (intermediate)

In chemistry, a chemical formula is far more than just a label; it is a precise quantitative recipe that tells us the exact ratio of elements within a substance. Take Magnesium Carbonate (MgCO₃) as our prime example. By looking at this formula, we can interpret that one formula unit contains one atom of Magnesium (Mg), one atom of Carbon (C), and three atoms of Oxygen (O). This shorthand notation allows scientists to communicate complex reactions concisely and ensures that the Law of Conservation of Mass is respected during chemical equations Science, Class X (NCERT 2025 ed.), Chemical Reactions and Equations, p.3. To move from identifying atoms to calculating physical mass (grams), we use the concept of Molar Mass. Every element has a specific atomic mass (expressed in unified mass units, 'u'). To find the total mass of a compound like MgCO₃, we sum the masses of all its constituent parts. Because the formula includes three oxygen atoms, we must multiply the atomic mass of oxygen by three before adding it to the masses of magnesium and carbon. This fixed proportion is what makes chemical formulas reliable for industrial and laboratory calculations—whether you are working with simple salts like Sodium Chloride or more complex ones like Magnesium Sulphate Science, Class X (NCERT 2025 ed.), Acids, Bases and Salts, p.28. Understanding these ratios is the key to Stoichiometry, the math behind chemistry. For instance, if you know that oxygen makes up a specific fraction of the total mass of MgCO₃, you can calculate exactly how much of the whole compound you need to extract a specific amount of oxygen. This logical leap—from the number of atoms in a formula to the weight of the substance in a beaker—is the foundation of all chemical engineering and pharmaceutical formulations.

Element	Atoms in MgCO₃	Contribution to Mass (Approx.)
Magnesium (Mg)	1	24.3 u
Carbon (C)	1	12.0 u
Oxygen (O)	3	3 × 16.0 = 48.0 u
Total (Molar Mass)	5 atoms	84.3 u

Sources: Science, Class X (NCERT 2025 ed.), Chemical Reactions and Equations, p.3; Science, Class X (NCERT 2025 ed.), Acids, Bases and Salts, p.28

6. Stoichiometry: Mass Percentage of Elements (exam-level)

At the heart of stoichiometry lies the principle of fixed chemical composition. As we understand from the nature of matter, compounds are substances where elements combine in a definite, unchanging ratio by mass (Science, Class VIII, Nature of Matter: Elements, Compounds, and Mixtures, p.131). This means whether you have a gram or a kilogram of a compound like Magnesium Carbonate (MgCO₃), the proportion of Magnesium, Carbon, and Oxygen within it remains constant. This stability is why we can use a chemical formula to represent a molecule—a stable particle formed when two or more atoms combine (Science, Class VIII, Nature of Matter: Elements, Compounds, and Mixtures, p.123).

To determine the Mass Percentage of an element in a compound, we compare the mass contributed by that specific element to the total molar mass of the entire compound. For instance, to find how much oxygen is in a molecule, we first identify the number of oxygen atoms present in the chemical formula (Science, Class X, Chemical Reactions and Equations, p.4). If a compound has three oxygen atoms, we multiply the atomic mass of oxygen (16.00 u) by three to get the total mass contribution of oxygen (48.00 g/mol) for every mole of the compound. The percentage is then calculated as:
Mass % of element = (Mass of element in 1 mole of compound / Molar mass of compound) × 100

This concept is incredibly powerful because it allows us to work backwards using the Unitary Method. If you know the total molar mass of a compound and the mass of one of its constituent elements, you can calculate the total mass of a sample required to provide a specific amount of that element. For example, if you know that one mole of a substance contains a certain mass of 'Element X', and you only need half that mass of 'Element X' for a reaction, you intuitively know you only need half a mole of the entire compound. This stoichiometric logic is the foundation for industrial chemical production and laboratory analysis alike.

Sources: Science, Class VIII, Nature of Matter: Elements, Compounds, and Mixtures, p.123; Science, Class VIII, Nature of Matter: Elements, Compounds, and Mixtures, p.131; Science, Class X, Chemical Reactions and Equations, p.4

7. Solving the Original PYQ (exam-level)

Now that you have mastered molar mass and mole-to-mass conversions, this question serves as the perfect application of stoichiometry. By looking at the chemical formula MgCO3, you apply the "building block" principle: the subscript '3' tells us there are three atoms of oxygen for every one formula unit of the compound. As outlined in NCERT Class 11 Chemistry, understanding the mass composition of an element within a compound is the key to bridging the gap between a molecular formula and a laboratory-measured quantity.

To solve this efficiently, first determine the mass of oxygen in exactly one mole of MgCO3. Since one mole of the compound contains three moles of oxygen atoms (3 × 16.00 g/mol), you have 48.00 g of oxygen for every 84.30 g of MgCO3. The question asks for the mass of the compound containing 24.00 g of oxygen—which you should immediately recognize as exactly half of the oxygen found in one full mole. Therefore, you simply need half a mole of the compound: 84.30 g / 2 = 42.15 g, making (A) the correct choice.

UPSC often sets "distractor" traps to catch students who skip steps or rush their logic. Option (B) 84.30 g is a classic trap for those who identify the total molar mass but fail to apply the stoichiometric ratio of oxygen. Option (C) 126.00 g might lure students who incorrectly multiply the mass or miscalculate the oxygen ratio. Success in these questions depends on your ability to recognize that chemical formulas represent definite proportions; once you see the 2:1 relationship between the standard molar oxygen (48 g) and the required amount (24 g), the calculation becomes an elegant, one-step mental exercise.