Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Uh-Oh! That’s all you get for now. We would love to personalise your learning journey. Sign Up to explore more. Sign Up or Login Skip for now Alternatively, you can relate the molar mass of the molecular formula and the empirical formula in the same way: Please do not block ads on this website. If carbon and hydrogen are present in a compound in a ratio of 1:2, the empirical formula for the compound is CH2. The molecular formula for the same compound will equal to n × (CH2), in other words, the molecular formula for this compound will be CnH2n Using the periodic table of the elements we can determine the empirical formula mass of this compound(CH2): 12.01 + (2 × 1.008) = 14.026 If we know the molecular mass of the compound is 28.00 (Mr = 28.00) then we can find the value of "n" in the molecular formula CnH2n: Subsitute this value, n = 2, back into the general molecular formula CnH2n to get the molecular formula of this compound. C(1 × 2)H(2 × 2) which is C2H4 There are many compounds that can have the empirical formula CH2 and therefore a molecular formula of the form CnH2n. Examples include:
Do you know this? Join AUS-e-TUTE! Play the game now! We can use the percentage composition (percent composition) of a compound to determine an empirical formula for the compound. Remember that the percentage composition gives us the percent by mass of each element present in the compound, for example If a compound is made up of 10% by mass of element X and 90% by mass of element Z On the other hand, the empirical formula of the compound gives us the lowest whole number ratio of atoms of each element in the compound, for example Empirical formula XZ tells us that for every 1 atom of X there is 1 atom of Z in the compound. Clearly, we will need to convert the "mass" of each element in the compound into a "number of atoms". 6.02 × 1023 atoms of element X = 1 mole of atoms of element X So we can use the "mass" and "molar mass" (or relative molecular mass) to calculate the moles of atoms of each element present,
Next we need to write the relationship between moles of X and moles of Z in the compound:
and then to find the lowest whole number ratio of these moles
Be aware that the calculation above to get the mole ratio of one element to another may not result in whole numbers. A further calculation may be needed to get the lowest whole number ratio of moles of one element to another. This is where a recognition of the decimal equivalent of common fractions, as shown in the table below, can be very helpful:
Once you have calculated the lowest whole number ratio of moles of one element to the other, you are ready to write the empirical formula, for example
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Question: A compound is found to contain 47.25% copper and 52.75% chlorine. Find the empirical formula for this compound. Solution:
Empirical formula for this compound is CuCl2
Can you apply this? Join AUS-e-TUTE! Take the exam now! Question: A compound with a molar mass of 34.0 g mol-1 is known to contain 5.88% hydrogen and 94.12% oxygen. Find the molecular formula for this compound. The solution to this problem is in two parts: Part 1: Determine the empirical formula using the percentage composition of the compound
Part 2: Determine the molecular formula using the empirical formula and molar mass of the compound
(1) Only a molecule, that is a substance made up of 2 or more atoms covalently bonded in a discrete unit, can have a molecular formula. This empirical formula of an ionic compound therefore tells us the lowest whole number ratio of cations to anions. (2) And yes, there are even more possibilities if you think about the structural isomers of butene such as: but-1-ene (1-butene) and but-2-ene (2-butene) |