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To find the percent composition of oxygen in Na2O, find the total molar mass of the compound. Then, divide the molar mass of oxygen by the molar mass of the compound, and multiply by 100% to get the percent oxygen.
mass % of element X = mass of element X ____________________ X 100 total mass of compound or mass of solute _____________________________ X 100 mass of solute + mass of solvent
For a chemical compound, ex. ABC: Percent (%) of A is (atomic weight of A x 100)/molecular weight of ABC.
it can be calculated using the formula percentage composition of N =Gram molecular mass of nitrogen in the compound/ Gram molecular mass of compound *100
A way of describing the concentration of solute in a compound or mixture is called Mass percentage or percent composition by mass. It can be calculated by dividing the mass of a component by the total mass of the mixture, then multiply the answer by 100%.
((mass of element)/(mass of compound))*100
To work out the mass of one element within a compound (ie the mass of Lithium in Lithium Carbonate), first work out the RFM (Relative Formula Mass) of the compound. To do this, look on the periodic table (PT) & add up the RAM (Relative Atomic Mass) of all of the atoms in the compound (the RAM is the larger of the 2 numbers in each box on the PT). So, if the formula for one molecule of Lithium Carbonate is Li2CO3, then the RFM is Li + Li + C + O + O + O = 7 + 7 + 12 + 16 + 16 + 16 = 74 Next work out what percentage of the compound is the element you want to know. So, Lithium accounts for 14 (2 Lithiums = 7 x 2) out of every 74 grams of Lithium Carbonate. As a percentage, this is ( 14 / 74 ) x 100 = 18.9 %. So, 18.9% of 1.55g [ ( 1.55 / 100 ) x 18.9 ] = 0.29g So there is 0.29g of Lithium is 1.55g of Lithium Carbonate.
To find the percent composition of oxygen in Na2O, find the total molar mass of the compound. Then, divide the molar mass of oxygen by the molar mass of the compound, and multiply by 100% to get the percent oxygen.
mass % of element X = mass of element X ____________________ X 100 total mass of compound or mass of solute _____________________________ X 100 mass of solute + mass of solvent
The atomic mass of Lithium is equal to the weighted average of the two naturally occurring isotopes. Let M=the atomc mass of Lithium; M1=the mass of lithium-6;M2=the mass of Lithium-7; %a=the abundance of lithium-6; %b=the abundance of lithium-7. M=(M1*%a+M2*%b)/100 we know that %a and %b must add up to give us 100%, so %a=100-%b. Therefore M=(M1*(100-%b)+M2*%b)/100 Then we just need to solve for %b because everything else is known. M=(M1*100-M1*%b+M2*%b)/100-->M-M1=(-M1*%b+M2*%b)/100 100(M-M1)=(-M1*%b+M2*%b)-->%b=100(M-M1)/(M2-M1)=92.5% Then %a=100-%b=100-92.5=7.5%
For a chemical compound, ex. ABC: Percent (%) of A is (atomic weight of A x 100)/molecular weight of ABC.
For a chemical compound, ex. ABC: Percent (%) of A is (atomic weight of A x 100)/molecular weight of ABC.
it can be calculated using the formula percentage composition of N =Gram molecular mass of nitrogen in the compound/ Gram molecular mass of compound *100
The formula shows that oxygen is one atom out three in the formula unit, so that the atomic percent of oxygen in the compound is 33.333.......... By adding the gram atomic masses of the three elements, one obtains 23.95 as the gram formula mass, to the justified number of significant digits (the four digits in the value for the gram atomic mass of lithium), and 100 (15.9994/23.95) is 66.80 percent by mass of oxygen in LiOH.
The molar mass of LiOH is 24,01874. The atomic weight of oxygen is 15,9994.24,01874 g----------------100 %15,9994 g---------------------xx = 15,9994 . 1oo/24,01874 = 66 %
Amount of mass of component Y divided by total mass times 100% equals (=) percent (%) Y component.
The molar mass is 20 g.