Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. Notice that the 2 on the right hand side is "distributed" to both the Na 2 and the O. Now the equation reads: 2Na + O 2 = 2Na 2O To fix this a 2 is added in front of the Na 2O on the right hand side. On the left hand side there are 2 O atoms and the right hand side only has one. In the next step the oxygen atoms are balanced as well.
In this there are 2 Na atoms on the left and 2 Na atoms on the right. This problem is solved by putting a 2 in front of the Na on the left hand side: 2Na + O 2 = Na 2O As it stands now, there is 1 Na on the left but 2 Na's on the right. In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. Hydrogen and oxygen are usually balanced last. Generally, it is best to balance the most complicated molecule first.
MOLECULAR EQUATION CALCULATOR TRIAL
Simple chemical equations can be balanced by inspection, that is, by trial and error. Using Trial and Error/Inspection Example #1 (Simple) By changing the scalar number for each molecular formula, the equation may be balanced. In case of net ionic reactions, the same charge must be present on both sides of the unbalanced equation. Thus, each side of the equation must represent the same quantity of any particular element. Assuming the acid reacts with all the iron(II) and not with the copper, how many grams of H 2(g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.In a chemical reaction, the quantity of each element does not change. He accidentally breaks off a 1.203 cm 3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. One liter of alloy completely fills a mold of volume 1000 cm 3. % error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%Įxample 10: Complex Stoichiometry ProblemĪn amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. % error = | theoretical - experimental | / theoretical * 100% Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass.Įxperimentally determined mass = 120.056 g/mol Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula.Ħ. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass.ĥ. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. Divide the experimentally determined molecular mass by the mass of the empirical formula. Determine the molecular mass experimentally.
In the example above, it was determined that the unknown molecule had an empirical formula of CH 2O.ġ. (1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol Oįrom this ratio, the empirical formula is calculated to be CH 2O. (0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol OĬonstruct a mole ratio for C, H, and O in the unknown and divide by the smallest number. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O 2 reactant.Ġ.333mol CO 2(44.0098g CO 2/ 1mol CO 2) = 1.466g CO 2ġ.466g CO 2 + 0.599g H 2O - 1.000g unknown organic = 1.065g O 2ġ.065g O 2( 1mol O 2/ 31.9988g O 2)( 2mol O/ 1mol O 2) = 0.0666mol O Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. This will give you the number of moles from both the unknown organic molecule and the O 2 so you must subtract the moles of oxygen transferred from the O 2.Ġ.0333mol CO 2 ( 2mol O/ 1mol CO 2) = 0.0666 mol OĠ.599g H 2O ( 1mol H 2O/18.01528 g H 2O)( 1mol O/ 1mol H 2O) = 0.0332 mol O Since all the moles of C and H in CO 2 and H 2O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample.Ġ.0333mol CO 2 ( 1mol C/ 1mol CO 2) = 0.0333mol C in unknownĠ.599g H 2O ( 1mol H 2O/ 18.01528g H 2O)( 2mol H/ 1mol H 2O) = 0.0665 mol H in unknownĬalculate the final moles of oxygen by taking the sum of the moles of oxygen in CO 2 and H 2O.
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