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What is the empirical formula of a compound that contains 20 percent calcium and 80 percent bromine by mass?
The empirical formula would be CaBr2 since it contains a ratio of 1 calcium to 2 bromine atoms.
What is the empirical formula of an oxide of Bromine which contains 71.4 percent Bromine by mass?
The empirical formula of the oxide of Bromine would be Br2O5. We can determine this by assuming a 100 g sample, which would contain 71.4 g of Bromine and 28.6 g of Oxygen. Then we convert these masses into moles and divide by the smaller value to find the mole ratio, which gives the empirical formula.
How do you know when to solve for a empirical formula?
You should solve for an empirical formula when you are given the percent composition of elements in a compound or when you have the molar mass of the compound but not the molecular formula. The empirical formula provides the simplest whole-number ratio of atoms in a compound.
What is the empirical formula of an iron oxide that 65.0 percent fe and 35.0 percent O?
The empirical formula of the iron oxide compound is Fe₂O₃. This is determined by dividing the percentage of each element by its molar mass to find the ratio of atoms in the compound.
What is the empirical formula of a compound that contains 53.73 percent Fe and 46.27 percent of S?
The empirical formula of the compound containing 53.73% Fe and 46.27% S is FeS. This is determined by dividing the given percentages by the molar mass of each element to find the mole ratio, which simplifies to FeS in this case.
Empirical formula of a compound containing 60.3 percent magnesium and 39.7 percent oxygen?
The empirical formula of this compound would be MgO.
What is the empirical formula of a compound that contains 20 percent calcium and 80 percent bromine by mass?
The empirical formula would be CaBr2 since it contains a ratio of 1 calcium to 2 bromine atoms.
What is the empirical formula of an oxide of Bromine which contains 71.4 percent Bromine by mass?
The empirical formula of the oxide of Bromine would be Br2O5. We can determine this by assuming a 100 g sample, which would contain 71.4 g of Bromine and 28.6 g of Oxygen. Then we convert these masses into moles and divide by the smaller value to find the mole ratio, which gives the empirical formula.
Calculate the empirical formula of a compound formed from 3 percent C 0.3 percent H and 96.7 percent you Which formula below correctly represents the empirical formula?
Chi a+
Calculate the empirical formula of a compound formed from 3 percent c 3 percent h and 96 7 percent you which formula below correctly represents the empirical formula?
CHI3
What is the empirical formula for a compound that is 7.9 percent Li and 92.1 percent Br?
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.
Can you determine the molecular formula of a substance from its percent composition?
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.
How do you know when to solve for a empirical formula?
You should solve for an empirical formula when you are given the percent composition of elements in a compound or when you have the molar mass of the compound but not the molecular formula. The empirical formula provides the simplest whole-number ratio of atoms in a compound.
What is the percent composition of a compound with the empirical formula CO2?
The percent composition of a compound with the empirical formula CO2 is 27.3% carbon and 72.7% oxygen.
What is the empirical formula of 22 percent Carbon 4.6 percent Hydrogen 73.4 percent Bromine?
To determine the empirical formula, we first need to convert the percentages to moles. Assuming we have 100g of the compound, we have 22g C, 4.6g H, and 73.4g Br. Next, we calculate the moles of each element: 22g C / 12.01 g/mol = 1.83 mol C, 4.6g H / 1.01 g/mol = 4.55 mol H, and 73.4g Br / 79.9 g/mol = 0.92 mol Br. Finally, we divide each mole value by the smallest mole value to get the simplest ratio, which gives us the empirical formula: C1H2Br1, or CH2Br.
What is the empirical formula for a compound that is 43.6 percent phosphorus and 56.4 percent oxygen?
p2o5
75.7 percent arsenic and 24.3 percent oxygen what empirical formula would it have?
As2O3
