Problem: What are the subscripts in the empirical formula of this compound?One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular kind of molecule by experimentally determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the molecule.For a molecule made up of elements A, B, and C, the proportions might be A:B:C2, meaning that there are two atoms of C for each atom of A and each atom of B. This may not be the actual description of the molecule (which might actually be A2B2C4), but it is the "reduced" version of that formula, called the empirical formula. The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula.A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound.

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What are the subscripts in the empirical formula of this compound?


One of the inherently satisfying features of chemistry is that chemical materials form and interact in a rational and predictable manner. For example, one can predict a great deal about a particular kind of molecule by experimentally determining the percentage composition of the elements in that compound. This gives us the relative proportions of the elements in the molecule.

For a molecule made up of elements A, B, and C, the proportions might be A:B:C2, meaning that there are two atoms of C for each atom of A and each atom of B. This may not be the actual description of the molecule (which might actually be A2B2C4), but it is the "reduced" version of that formula, called the empirical formula. The actual formula is some multiple of the empirical formula. To know the actual formula we need to know both the empirical formula and the molecular mass of the compound. This provides us with the multiplier value in whole units that must be applied to the empirical formula to get the actual formula.

A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound.

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