By determining the molecular mass, then dividing the molecular mass by the formula mass of the empirical formula to determine by what integer the subscripts in the empirical formula must be multiplied to produce the molecular formula with the experimentally determined molecular mass.
The information about the actual molar mass is superfluous. Given any molecular formula, the corresponding empirical may be obtained by dividing all the subscripts in the molecular by the largest integer that yield an integer quotient for each subscript. In the given formula, the empirical formula is CH2.
To find the molecular formula from the empirical formula, we need to know the molar mass of the empirical formula. In this case, the empirical formula's molar mass is 88. To find the molecular formula, we divide the given molecular mass (176) by the empirical formula's molar mass (88) to get 2. This means the molecular formula of Vitamin C is twice the empirical formula, so the molecular formula is C6H8O6.
To find the molecular formula from the empirical formula (C3H5O) and molar mass, you need to calculate the molar mass of the empirical formula. Then, divide the molar mass of the unknown compound by the molar mass of the empirical formula to get a ratio. Finally, multiply the subscripts in the empirical formula (C3H5O) by this ratio to determine the molecular formula of the unknown compound.
The empirical formula of C2H5 corresponds to an empirical mass of 29 g/mol. To find the molecular formula from the empirical formula and molecular mass, divide the molecular mass by the empirical mass to get the "scaling factor" (58 g/mol ÷ 29 g/mol = 2). Multiply the subscripts in the empirical formula by the scaling factor to get the molecular formula: C2H5 x 2 = C4H10. So, the molecular formula is C4H10.
In order to answer this question you need to know the molar mass of dinitrogen trisulfide (N2S3), and that 1 mole of molecules is equal to 6.022 x 1023 molecules. Molar mass is determined by multiplying each element's subscript by that element's atomic weight on the periodic table, and expressing it in grams/mole.1 mole N2S3 molecules = 6.022 x 1023 molecules N2S3molar mass N2S3 = 124.208g/molConvert molecules to moles.2.26 x 1025 molecules N2S3 x (1mol N2S3/6.022 x 1023 molecules N2S3) = 37.5 moles N2S3Convert moles to mass in grams.37.5mol N2S3 x (124.208g N2S3/1mol N2S3) = *4660 grams N2S3*The answer is rounded to three significant figures.
The actual mass must be divided by the empirical mass. This was derived from the following equation: (subscript)(empirical formula) = (molecular formula) subscript = (molecular formula)/(empirical formula)
The density or some other information must be given that allow you to find the molar mass. Calculate the empirical formula mass. Divide molar mass by empirical formula mass. This answer is multiplied by all subscripts of the empirical formula to get the molecular formula.
molar mass of unknown/molar mass of empirial = # of empirical units in the molecular formula. Example: empirical formula is CH2O with a molar mass of 30. If the molar mass of the unknown is 180, then 180/30 = 6 and molecular formula will be C6H12O6
Chemical Spider mention the name 2λ4-Diazathia-1,2-diene-1,3-diyldisulfanide.
molar mass over grams of elementThe above answer is somewhat correct. In order to find the molecular formula when given the empirical formula, you must first find the molar mass of the empirical formula.MOLAR MASS# atoms element A x Atomic Mass element A (Periodic Table) = mass A# atoms element B x atomic mass element B (periodic table) = mass B... etc.Add up all of the mass values found above and you have the molar mass.Then, after you have found the empirical formula's molar mass, you divide the molar mass of the molecular formula by the empirical formula's molar mass (solving for n).MOLECULAR FORMULA EQUATION: N (Empirical formula) (read as N times empirical formula) where:N = Molar mass substance---- Molar Mass emp. form.
molar mass over grams of elementThe above answer is somewhat correct. In order to find the molecular formula when given the empirical formula, you must first find the molar mass of the empirical formula.MOLAR MASS# atoms element A x Atomic Mass element A (Periodic Table) = mass A# atoms element B x atomic mass element B (periodic table) = mass B... etc.Add up all of the mass values found above and you have the molar mass.Then, after you have found the empirical formula's molar mass, you divide the molar mass of the molecular formula by the empirical formula's molar mass (solving for n).MOLECULAR FORMULA EQUATION: N (Empirical formula) (read as N times empirical formula) where:N = Molar mass substance---- Molar Mass emp. form.
To determine the molecular formula from the empirical formula and gram formula mass, first calculate the empirical formula mass of C4H9 (4 carbons + 9 hydrogens). Then, divide the gram formula mass by the empirical formula mass to find the ratio. Finally, multiply the subscripts in the empirical formula by this ratio to get the molecular formula, which in this case is C8H18.
By determining the molecular mass, then dividing the molecular mass by the formula mass of the empirical formula to determine by what integer the subscripts in the empirical formula must be multiplied to produce the molecular formula with the experimentally determined molecular mass.
N2S3 is properly named dinitrogen trisulfide. This name reflects the composition of the compound, with the ratio of two nitrogen atoms to three sulfur atoms.
the definition of 'empirical mass' is the simplest ratio of the molecular mass.for example: glucose.the molecular mass of glucose is C6 H12 O6 = 180but the empirical mass of glucose is C H2 O = 30hope this helps. it's all about ratios.
To determine the molecular formula from the empirical formula and the vapor density, we first need to calculate the empirical formula mass. The empirical formula mass of CH2O is 12g (carbon) + 2g (hydrogen) + 16g (oxygen) = 30g/mol. Next, we compare the vapor density with the empirical formula mass to find the factor by which the empirical formula mass is multiplied to get the molecular mass. The factor is 1.5 (45 / 30 = 1.5). Finally, we multiply the subscripts in the empirical formula by this factor to get the molecular formula, which is C3H6O3.