The Emp ical Formula of an Oxide oxide ned in a che the change in mass after a r

Question

The Emp ical Formula of an Oxide oxide ned in a che the change in mass after a reactio can solve reaction. Using data, we Background Information: Today, we are going to in several antacids. You will need to after all, people are going to eat this stuff!). assume that you work for r a pharmaceutical company. You are oxide, an active ingredient eaction and get rid of the impurities combine with them in different ow elements can combine with different e even ht) or Ihe empirical formula is the simplest ratio of atoms inyo The moleculade to calculate ratios. A with different elements and even N2O (an old ao s). These three molecules have extremely different traits, the mirogen and oxygen. In this reaction, we will determine example is the NOi the addition of nitrogen with oxygen. These two ), or NO2 (a poisonous three they are made only of nitrogen and ght (h is used by our bodies to 'talk' to the brain), N you c the small whole numbers that accurately express empirical forahe molaf mass formula molecular formula, we need the molar obers that smplest ratio of atoms involved in a combination. It is made of formula mass empirical we will get mass of our compound and the molaf mass of the cal formula. a. By dividing the molar mass of our compound by o the empirical foul e the number of units contained in the true formula by multiplying t n ere me we heated is an example set of calculations using sulfur combining with oxygen. Assu of sulfur in an open crucible. After reacting with oxygen in the air, we rewe product and found it weighted 5.20 g. Where did this extra mass come from? Did we create ti sufur? Nope, the sulfur reacted with oxygen in the air. We can figure out how much oxygen is Mass of Oxygen-5.20 g-2.60 g 2.60 gO Next we can convert each of those masses into moles. We do this by dividing by the molar mass. Moles of O Mass of O/ Molar mass of O: 2.60g O/ (16.00g/mol) 0.163 moles O Moles of S - Mass of S/ Molar mass of S: 2.60g O/(32.07g/mol)-0.0811 moles S our product by subtraction. Now that we know how many moles of each are present in our product, we will divide each number by the smallest amount of moles. In this case, S has the fewer number of moles. This

Explanation / Answer

As the mass of Mg ribbon is not given,let us use arbitrary amounts of masses to see through the calculations:

Mass of Mg ribbon=1.5g

Rxn taking place,

Mg reacts O2 to give magnesium oxide with formula MgxOy

mass of product (MgO)=2.518g

Mass of O=Mass of product -mass of Mg=2.518g-1.5g=1.019g

mol of Mg=mass of Mg/molar mass of Mg=(1.5g/24.305g/mol)=0.0617 mol

mol of O=mass of O/molar mass of O=1.019g/16.0g/mol=0.0637 mol

ratio of Mg to O=0.0617:0.0637=0.968:1=1:1

Emperical formula: MgO

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