Chemistry

Elements and Compounds

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The molecular formula of this compound is N2H2. This is obvious because the empirical formula of a compound is the lowest positive integer ratio of atoms present.

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0Calculate the empirical formula weight. Find the ratio of the molecular weight to the empirical formula weight. (n= molecular weight/ empirical formular weight). Multiply each subscript of the empirical formula by n.

If you mean to find its molecular formula: 1. First you must obtain the empirical formula. Find the molar mass of the empirical formula. 2. The molar mass of the entire molecule must be given or its molecular formula cannot be find. 3. Put the molar mass given over the molar mass of the empirical formula. 4. Use this number (kind of like a scaler) to multiply by the entire empirical formula. This is the molecular formula.

Because an empirical formula is the simplest form of a compound, we know that the molecular formula contains more atoms than it does. Since we are given the molar mass, we can use this formula. x ( MM of empirical formula ) = MM of molecular formula MM of empirical formula = 12(2) + 1(6) + 16 = 46 MM of molecular formula = 138 46x = 138 x= 138 / 46 x=3 Therefore, the molecular formula is 3(C2H6O) that is C6H18O3

You have to find its empirical formula using the percentage composition. When you have done that, work out the relative molecular mass (Mr) of the empirical formula. This should be a multiple of the compound's Mr, so you multiply the amount of each atom in the empirical formula by this number, which gives you the final molecular formula.

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.

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.

There are 4 step to determine molecular formula, which are given bellow Step:1:- Find empirical formula Step:2:- Find empirical formula mass Step:3:- Find n n=molecular mass/empirical formula mass Step:4:- now find molecular formula to find molecular formula molecular formula(empirical formula)n

The empirical formula of SN has a formula unit mass of the sum of the gram atomic masses of nitrogen and sulfur, i.e., about 46.0667. The gram molecular mass given in the problem divided by this formula unit mass is about 4. Therefore, the molecular formula is S4N4.

The empirical formula is representative for the chemical composition of a compound; the structural formula is representative for the spatial structure of the compound.

The gram formula unit mass of the empirical formula C2H3 is twice the gram atomic mass of carbon plus three times the gram atomic mass of hydrogen, or about 27. The nearest integer to 162.27/27 is 6. Therefore, the molecular formula for the compound is C12H18.

The chemical formula in which the subscripts are given in the smallest ratio.

The formula given for butane is a molecular formula. The empirical formula for any molecular is obtained by dividing all the element subscript numbers by the highest common factor of all the subscripts; in this particular instance leading to C2H5 as he empirical formula.

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.

The molecular mass is C2H7 you have 186.5 g of the substance, so calculate 2(12.01)+7(1.01)=31.09 so 186.5/31.09=6 so you can say C12H42 but when you divide 2 by 2 and 7 by 2 you obtain 1 and 3.5 because you have 3.5 you have to multiplied by 2 again, so you obtain the same formula as the empirical formula. It example is the same as water, which empirical formula es H2O as its molecular formula

In order to find a substance's molecular formula, proceed with the following steps. An element's molecular formula is either going to be equal to the empirical formula or is a multiple of the empirical formula (n).1. Find the empirical formula's molar mass. Calculate the molar mass of the empirical formula.# atoms element A x atomic mass element A = mass A# atoms element B x atomic mass element B = mass B... etc.Add up all of the mass values found above for the empirical formula's molar mass (MM).2. Solve this equation (below).n = MM compound (given) = Empirical Formula Units---- MM emp. form.3. After solving for n, multiply the empirical formula.Molecular Formula = N (Empirical Formula)Ex. What is the molecular formula of a compound which has an empirical formula of CH2 and a molar mass of 126.2g?1. Find the empirical formula's molar mass.1 atom carbon (C) x 12.01g = 12.01g2 atoms hydrogen (H) = 2.016gEmpirical Molar Mass = 14.032. Solve this equation (below).n = 126.2g substance = 9 Empirical Formula Units (n)--------- 14.03g CH23. After solving for n, multiply the empirical formula.Molecular Formula = 9 (CH2) --> C9H18

The formula as written, including a capital "I", is already an empirical formula, since it shows a total of 7 carbon atoms and none of the other subscripts is integrally divisible by 7. However, the formula may have been intended to be written C16H12Cl2O2. In that instance, all the subscripts can be integrally divided by 2 to given an empirical formula of C8H6ClO.

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.

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.

depends, an Empirical formula will always (by definition) show the ratio in which atoms are combined within a molecule. a molecular formula on the other hand shows the number of atoms of each element in a molecule, the only exceptions being massive structures, crystal latices and where a molecular formula cannot be produced and is either given in the form n*(empirical formula) or just the empirical formula. do note however that reference to a formula of a molecule usually refer to the molecular formula except in the exceptions listed above.

The empirical formula should have the lowest subscript numbers possible, consistent with the actual proportions of atoms present. In the formula given, all the subscripts are divisible by 4; therefore, the empirical formula should be C4H8O2N.

If you are given the empirical formula and are asked for the actual formula, then the molecular mass of the compound will be given too. Take this example problem: Empirical Formula: CH2O Molecular Mass: 180.0 First you have to find the empirical mass. Just find the atomic masses of all the elements in the empirical formula and add them together. If there are multiple atoms of the same element, then you have to add the element's atomic mass for every multiple. In the example, you have to add hydrogen twice because there are two hydrogen atoms in the empirical formula. C- 12.0 H- 1.00 H- 1.00 O- 16.0 + _________ 30.0 The empirical formula is some multiple of the actual formula. The empirical formula shows the ratio of atoms as 1:2:1. This means that the actual formula could be 2:4:2 or 3:6:3 or 4:8:4 etc. In order to find what multiple it is, divide the molecular mass by the empirical mass. An easier way to think of it is: x(empirical mass) = molecular mass Use this formula to find x: x(30.0) = 180.0 x = 6 In this example, x turned out to be exactly 6, but in some cases (especially in lab results) the answer will be close to a whole number but not exact. There is usually a standard +/- .02 for your result. If the x value was 6.02 or 5.98, we would just round up or down as long as it is within .02 Take the x value and multiply it with the number of atoms in the original empirical formula. C1H2O1 * 6 = C6H12C6 The final result is the actual molecular formula.

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 requirements for an empirical formula are that it give the correct ratios between all pairs of atoms in the actual molecule and have subscripts (including the value 1 implied by lack of an explicit subscript) with the lowest possible values to achieve all the correct ratios. To meet the latter condition, either one subscript must be one or the two smallest subscripts must be distinct prime number. The formula of the compound given has only one prime number subscript, 5. If all subscripts in the molecular formula are divided by the same number to yield an integral quotient, the resulting formula will still be empirically correct. In the given molecular formula, all the subscripts are integrally divisible by 5. Therefore, the empirical formula is CH3O2.

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.

An empirical formula may be obtained from a molecular formula by dividing all of the subscripts in the formula by every common integral factor of all the subscripts until reaching a collection of subscript numbers that have no common integral factor except the number 1. In the given instance, the only common integral factor is 2; therefore, the empirical formula is CH3O.

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