Chemistry
Elements and Compounds

# What is the empirical formula of a compound containing 36.48 percent Na?

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Impossible to answer without knowing the other element(s).

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## Related Questions

Cr3Si2 is the empirical formula for a compound containing chromium and silicon an has 73.52 mass percent chromium.

The empirical formula for a molecule containing 46.8 percent si and 53.2 percent o is in the middle of the two. These added together and divided by two will net the average result.

If it tells you to find the empirical formula when percent composition is given or if the mass of each element is given in a specific compound.

Percent composition can be used to calculate the percentage of an element/compound in a mixture. From the percent composition, you can also find the empirical formula. And from the empirical formula you can find the actual molecular weight.

The empirical formula is the simplest ratio of the elements within a compound. Therefore, it can be used to calculate the percentage of an element within a compound. For example, the empirical formula for sodium chloride is NaCl. From this, we can see that the ratio of sodium ions to chloride ions is 1Na : 1Cl. Therefore, a sodium chloride molecule is composed of 50% sodium and 50% chloride.

First find empirical formula by assuming you have 100 grams of the compound -85.6 grams of Carbon and 14.4 grams of Hydrogen Divide both these amounts by their mass per mole. Carbon: 85.6/12 = 7.13 Hydrogen: 14.4/1.008 = 14.28 Now divide the larger answer by the smaller answer. 14.28 divided by 7.13 to get a ratio of 2 H for every one Carbon in the empirical formula Empirical formula = CH2

Divide each percentage by the atomic mass of that element. Nitrogen: 30.4 / 14 = 2.1714 Oxygen: 69.6 / 16 = 4.35 compare these and divide by the smallest, round if necessary. Nitrogen: 2.1714 / 2.1714 = 1 Oxygen: 4.35 / 2.1714 round to 2 This gives an empirical formula of NO2 The mass of this empirical formula = 14 + 2x16 = 46 Divide the molar mass by this empirical mass. 92.0 / 46 = 2 multiply the subscripts in the empirical formula by this number to get the molecular formula (NO2)2 = N2O4 THIS IS THE MOLECULAR FORMULA.

This compound is the trioxygen diarsenic oxide (As2O3), containing 75,7442 % As and 24,2558 % oxygen.

Not completely. The empirical formula of a substance can be determined from its percent composition, but a determination of molecular weight is needed to decide which multiple of the empirical formula represents the molecular formula.

CH2 is the empirical formula of a substance that consists of 85.60% carbon and 14.4% hydrogen.

that is empirical formula but percent mass of each element is:Carbon ==> 60.00%Hydrogen ==> 4.44%Oxygen ==> 35.56%

Based on % composition, one can determine the moles of each element in, say, 100 grams of compound. Then, one can see the mole ratio of all the elements in the compound, and adjust them so as to obtain whole numbers in the lowest possible ratio. This is then the empirical formula.

We can't tell. What's the other 90 percent? If you meant 40/60 instead... the mass of sulfur is twice that of oxygen, so a mass ratio of 40:60 is equivalent to an atom ratio of 1:3. The empirical formula would be SO3.

Because unlike the empirical formula, the molecular formula does not have to be the simplest ratio.If by chance you are given the percent composition of the elements in a substance, you could calculate the empirical formula and then the empirical formula's mass. However, the molecular formula equation is molecular formula= (empirical formula)n, where n is the mass of the molecular formula divided by the mass of the empirical formula. You would, therefore, need to know the mass belonging to the molecular formula, which you are not given.

You can only calculate the empirical formula because you do not have a mass of this compound given. To do the empirical formula assume 100 grams and change percent to grams. Get moles. 80 grams Carbon (1 mole C/12.01 grams) = 6.66 moles C 20 grams hydrogen (1 mole H/1.008 grams) = 19.84 moles H the smallest becomes 1 in the empirical formula and the other number is divided by it, Thus; H/C 19.84 moles H/6.66 moles C = 2.9, which we call 3 so, CH3 --------------- is the empirical formula To get the molecular formula tour question needed to read; How to calculate molecular formula from such ans such mass of compound with these percentages of elements, Which, of course, your question did not provide. Then you would have divided that given mass by the mass total of the elements of the empirical formula, got a whole number by which you would have multiplied the numbers of your empirical formula to get molecular formula.

The atomic mass of lithium is 6.94, while that for bromine is 79.90. 7.9/6.94 = 1.095 and 92.1/79.90 = 1.15. the closest integer ratio between these two number si 1:1, so the empirical formula is LiBr.

This compound is UF6. Atomic weight of uranium: 238,02891 Atomic weight of fluorine: 18,9984032 Molecular mass of UF6: ca. 352,04121

Firstly divide percentages by molar mass of that elementNa43.4/23=1.9C11.3/12.01=0.94O45.3/16=2.83Then divide the result of the first step by the smallest answerNa1.9/.94=2C.94/.94=1O2.83/.94=3Empirical Formula is Na2CO3

You can find the empirical formula of a compound using percent composition data. If you know the total molar mass of the compound, the molecular formula usually can be determined as well. The easiest way to find the formula is:Assume you have 100 g of the substance (makes the math easier because everything is a straight percent).Consider the amounts you are given as being in units of grams.Convert the grams to moles for each element.Find the smallest whole number ratio of moles for each element

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