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What is the Empirical formula of the compound that is 52.7 percent K 47.3 percent Cl?
The empirical formula of the compound with 52.7% K and 47.3% Cl is KCl (potassium chloride). This is because the ratio of potassium to chlorine atoms in the compound is 1:1, leading to the simple formula KCl.
What is the empirical formula of a compound that contains 36.1 ca and 63.9 cl?
To find the empirical formula, we first convert the masses of each element to moles. For calcium (Ca), with a molar mass of approximately 40.1 g/mol, 36.1 g of Ca corresponds to about 0.90 moles. For chlorine (Cl), with a molar mass of approximately 35.5 g/mol, 63.9 g of Cl corresponds to about 1.80 moles. The ratio of moles of Ca to Cl is 0.90:1.80, which simplifies to 1:2. Thus, the empirical formula of the compound is CaCl₂.
What is the empirical formula of a molecule containing 24 carbon (C) 6 hydrogen (H) and 70 chlorine (Cl)?
To find the empirical formula, we first determine the simplest whole-number ratio of the elements. The given molecule has 24 carbon atoms, 6 hydrogen atoms, and 70 chlorine atoms. Dividing each by the greatest common divisor, which is 2, we get the ratio of C: 12, H: 3, Cl: 35. Therefore, the empirical formula is C₁₂H₃Cl₃₅.
What is the emperical formula of a sample that has 80g Ca and 140g Cl?
The molar mass of Ca is 40 g/mol, and the molar mass of Cl is 35.5 g/mol. To find the empirical formula, we must first convert the masses to moles. 80 g Ca is 2 moles and 140 g Cl is 4 moles. To get the simplest whole number ratio, we divide by the smallest number of moles, giving us a ratio of 1 Ca to 2 Cl atoms, and the empirical formula is CaCl2.
What is the empirical formula for magnesium oxide?
MgCl2 is the correct formula, or two atoms of Cl for each 1 atom of Mg. You could also say 2 moles of Cl atoms for every 1 mole of Mg atoms. The empirical formula should be written with the subscripts in lowest whole number terms. Note that this formula is consistent with the fact that an Mg cation is Mg+2 and a Cl anion is Cl-1. Given these charges, a formula of MgCl2 is the one with the smallest whole number subscripts that will allow the charge of the compound to be zero. See related question below for more details on how to find empirical formulas.
If 1.164g of iron fillings reacts with chlorine gas to give 3.384 g of iron chloride what is the empirical formula of the compound Use the mole ratio method.?
To find the empirical formula, calculate the moles of iron and chlorine in the compound. Then, determine the ratio of moles of iron to moles of chlorine. The mole ratio is 1:2, so the empirical formula is FeCl2.
What is empirical formula of chlorine and Iodine?
Chlorine and iodine are both elements that form diatomic molecules (molecules that consist of two atoms). One atom of chlorine is symbolised by Cl; however, single atoms of chlorine do not exist unattached to other atoms. One molecule of chlorine is Cl2. The situation is the somewhat similar for iodine, and a molecule of iodine is I2. Please see the links.
What a possible formula for the empirical formula of C3H5ClO?
To find the empirical formula of a compound, you need to simplify the ratio of its atoms to the smallest whole numbers. For C3H5ClO, the subscripts are 3 carbon (C), 5 hydrogen (H), 1 chlorine (Cl), and 1 oxygen (O). The greatest common divisor of these numbers is 1, so the empirical formula is the same as the molecular formula: C3H5ClO.
What is the empirical formula of KCl?
Potassium chloride is a ionic lattice which is made out of potassium ions and chloride ions. Therefore, separate 'molecules of this compound do not exist. The empirical formula itself is KCl which is used also as the chemical formula for the compound.
What is the empirical formula of C10H4?
It is an empirical formula.
empirical formula of a 9.50g sample of a compound that contains 3.03g of K 2.75g Cl and O?
This compound is potassium oxide, K2O.
What is the empirical formula of a compound that contains 75 percent Ag and 25 percent Cl by mass?
The empirical formula of the compound would be AgCl, as the ratio of silver to chlorine in the compound is 3:1 based on the given mass percentages (75% Ag and 25% Cl). This ratio simplifies to AgCl when expressed in the simplest whole number ratio.
