Related Questions

The molecular mass of a compound with the formula CH2O is approx. 30, not 120.

Find the molar mass of CH2O, which is roughly 30amus Then divide the molecular mass of 180.15amus by 30 = 6 Multiply each element by 6 to give C6H12O6

a) molecular formula b) molecular weight c) percentage ratio d) number of atoms

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.

No. It must be a whole number. Since the empirical formula of a compound shows the proportions of the elements in the simplest whole number ratio there is going to be at least one odd number in the formula. Multiplying by 2.5 would then result in you having half an atom somewhere in the molecule, which you can't really have.

find empirical formula mass which is 16g.then divide 32g with 16g.the result which is 2, multiply it with the emf to get the molecular formula.answer = n2h4.

First lets assume you mean percent by mass which are the usual numbers given. The determination of chemical formula from percentage by weight is achieved through the use of the concepts of molecular mass, empirical formula, and molar mass. The molecular mass of a compound can be found by taking its constitute atoms and adding their relative atomic mass (a number given in atomic mass units (u), where a carbon 12 isotope is defined as having a mass of 12). This will give a compound's molecular mass for example - carbon dioxide is composed of one carbon atom, and two oxygen atoms; the relative atomic masses being to three significant figures 12.0 and 16.0 respectively to three significant figures; by adding 12.0 to 16.0 x 2 you find 48.0 the mass of a carbon dioxide molecule. A compounds empirical formula is the chemical formula expressed in the simplest whole number ratio. For example take sucrose C12H22O11 the empirical formula of which being: C1H2O1. The molar mass of a compound is a mass in grams equal to the relative atomic mass, that is if you take our carbon dioxide molecule above with a molecular mass of 48.0 its molar mass is 48.0grams. Now, given a compounds percent composition and molecular mass you can find it's chemical formula. The first step in this process is finding the empirical formula. To find the empirical formula given percentages, as a convenience image you had a 100 gram sample. In this imagined sample you need to take the masses of each element you would have (it's the same as the percent weight) and divide this by the molar mass to find moles. Next having done this find the simplest whole number ratio between moles of each element. This will give you the empirical formula. Now having found the empirical formula we need to find the molecular mass of such a compound if it did exist. The molecular mass is found as above by multiplying each of the compound constitute parts by its relative atomic mass. With the empirical formula its molecular mass, and the molecular mass of the actually compound in hand we can calculate the chemical formula. Take the molecular mass for the compound and divide this by the molecular mass of the empirical formula. This will give you a positive integer, or close enough to. Take this number and multiply each of the subscripts for the empirical formula by it. You are done. For example - A you have determined through analytical analysis that the percent composition by mass of a compound is C-42%, H-6.4%, and O-51.6% you know the compound has a molecular mass of approximately 342u, what is the chemical formula? First our 100g sample -42 grams Carbon -6.4 grams Hydrogen -51.6 grams Oxygen m(Carbon)/M(Carbon) = 42/12 = 3.5 m(Hydrogen)/M(Hydrogen) = 6.4/1 = 6.4 m(Oxygen)/M(Oxygen) = 51.6/16 = 3.225 C:H:O ≈ 1:2:1 Therefore the empirical formula of our compound is C1H2O1 M(C1H2O1) = 12.0 + 2 * 1.0 + 16.0 = 30.0u M(compound)/M(C1H2O1) = 342 / 30.0 ≈ 11 Therefore our compound is: C1*11H2*11O1*11which gives C12H22O11

The empirical formula for a compound is the collection of atomic symbols, each having a subscript (including possibly the implicit subscript "1" when no explicit subscript is shown), the collection of subscripts satisfying the following two conditions: The ratio of subscripts on any two distinct atomic symbols in the empirical formula is the same as the atomic ratio of the corresponding elements in the compound; and There is no integer except 1 that will divide all the subscripts in the empirical formula to yield an integer as quotient.

The compounds having two elements are the binary compounds as H2O, HF, HCl, HBr, HI.

Methane is an elementary organic compound having the molecular formula of CH4. Methane is composed of carbon (C) and hydrogen (H) atoms. The molecular weight of CH4 is 16. 01 grams per mole.

Two or more compound having same molecular formula but different structural Formulas are called isomers and process is called Isomerism.

A molecular formula shows how many of what kind of atoms are in a molecule of a compound. Water, having two atoms of hydrogen and one atom of oxygen has the molecular formula H2O. Sulfuric acid, with two hydrogen atoms, one sulfur atom, and four oxygen atoms, is written H2SO4.

== == For some reason, my calculations came up with the empirical formula for sucrose being COH2. Wiki says the molecular formula is something much larger as expected. Basically, all sugars have the ratio of C:H:O of 1:2:1, and as a result, all have the same empirical formula . That being said, any sugar can be a multiple of that ratio (except having non-integer numbers of atoms - .5 atoms) so sucrose (C11H22O11)/11 = that magic ratio of 1:2:1.

A compound is not a reaction; a compound is a molecule having a chemical formula.

Permutit is Sodium aluminosilicate having the formula Na2O.Al2O3.Si2O4.xH2O

Structural isomers have same molecular formula but different structural formula. Structural isomers for Butane having formula C4H10 are two. One is n-butane and the other is iso-butane.

The same way you calculate the molecular mass of any compound. Alternatively, if you know the mass of one isomer, you don't need to do any calculations. All isomers having the same formula have the same mass.

different structures all having same molecular formula

Compounds having this molecular formula are organic compounds.

C4H10 is the simplest alkane formula C2H4 is the simplest alkene formula

Sodium Nitrite is a reducing agent having the molecular formula,NaNO2.

These formulae cannot represent substances having the same empirical formulae because the numbers of atoms of each element in the two formulae are different.

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