Question

question 2 of 39
element x forms three different compounds with element y. information about these compounds is contained in the table.

compound	formula	mass of y per gram of x
2	xy?	1.41 g
3	xy?	0.94 g

what are the formulas of compounds 2 and 3? replace each question mark with the appropriate integer.

compound 2: \boxed{xy?} quad compound 3: \boxed{xy?}

Explanation:

Step1: Analyze the law of multiple proportions

The law of multiple proportions states that when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers. Let's take the fixed mass of X as 1 gram. For compound 1 (\(XY_6\)), mass of Y per gram of X is 2.82 g. Let the mass of Y per gram of X in compound 2 be \(m_2 = 1.41\) g and in compound 3 be \(m_3=0.94\) g.

Step2: Find the ratio of masses of Y

First, find the ratio of \(m_2\) to \(m_1\) and \(m_3\) to \(m_1\).

Ratio for compound 2: \(\frac{m_2}{m_1}=\frac{1.41}{2.82}=\frac{1}{2}\)

Ratio for compound 3: \(\frac{m_3}{m_1}=\frac{0.94}{2.82}=\frac{1}{3}\) (approximately, since \(2.82\div3 = 0.94\))

Since in compound 1, the formula is \(XY_6\) (6 atoms of Y per atom of X), for compound 2, since the mass of Y is half of that in compound 1, the number of Y atoms should be half of 6? Wait, no. Wait, the law of multiple proportions: the ratio of the number of atoms of Y in the compounds should be in the ratio of the masses of Y (since mass is proportional to number of atoms for the same element).

Wait, let's think again. Let the mass of X be fixed (say 1 mole of X, or 1 gram of X). For compound 1: \(XY_6\), so moles of Y per mole of X is 6. Mass of Y per gram of X is 2.82 g. Let the molar mass of Y be \(M_Y\) and molar mass of X be \(M_X\). Then, for compound 1: \(\frac{6M_Y}{M_X}=2.82\) (since mass of Y per gram of X is \(\frac{mass\ of\ Y}{mass\ of\ X}=\frac{6M_Y}{M_X}\) when we take 1 mole of X, mass of X is \(M_X\) grams, mass of Y is \(6M_Y\) grams, so mass of Y per gram of X is \(\frac{6M_Y}{M_X}\)).

For compound 2: let the formula be \(XY_n\), then \(\frac{nM_Y}{M_X}=1.41\)

Divide the equation for compound 2 by the equation for compound 1: \(\frac{\frac{nM_Y}{M_X}}{\frac{6M_Y}{M_X}}=\frac{1.41}{2.82}\)

Simplify: \(\frac{n}{6}=\frac{1}{2}\) (since \(\frac{1.41}{2.82}=0.5=\frac{1}{2}\))

So \(n = 6\times\frac{1}{2}=3\). Wait, that can't be. Wait, maybe I mixed up. Wait, the mass of Y per gram of X: for compound 1, it's 2.82 g Y per 1 g X. For compound 2, 1.41 g Y per 1 g X. So the ratio of Y in compound 2 to compound 1 is 1.41/2.82 = 1/2. So if in compound 1, the formula is \(XY_6\) (6 Y per X), then in compound 2, since Y is half, the number of Y should be 6*(1/2)=3? Wait, no, maybe the other way. Wait, maybe the fixed mass is of Y? No, the table says "Mass of Y per gram of X". So per gram of X, how much Y. So for compound 1: 1 g X combines with 2.82 g Y. Compound 2: 1 g X combines with 1.41 g Y. Compound 3: 1 g X combines with 0.94 g Y.

So the ratio of Y masses (compound 2 : compound 1) is 1.41/2.82 = 1/2. Ratio (compound 3 : compound 1) is 0.94/2.82 ≈ 1/3 (since 2.82 ÷ 3 = 0.94).

Now, according to the law of multiple proportions, the ratios of the number of atoms of Y in the compounds should be in the ratio of the masses of Y (since mass is proportional to number of atoms for the same element). So if compound 1 has 6 Y atoms per X atom, then compound 2 should have 6*(1/2)=3 Y atoms? Wait, no, that would be if the mass ratio is inverse. Wait, no, let's use moles. Let's assume we have 1 gram of X. Let the molar mass of X be \(M_X\), so moles of X is \(1/M_X\). In compound 1, moles of Y is (2.82 g)/\(M_Y\). The formula is \(XY_6\), so moles of Y = 6 * moles of X. So (2.82)/\(M_Y\) = 6*(1/\(M_X\)) → \(2.82 M_X = 6 M_Y\) → \(M_Y = (2.82 / 6) M_X\)

For compound 2: moles of Y = (1.41)/\(M_Y\), moles of X = 1/\(M_X\). Let the formula be \(XY_n\), so moles of Y = n * moles of X. So (1.41)/\(M_Y\) = n*(1/\(M_X\)) → \(1.41 M_X = n M_Y\)

Substitute \(M_Y\) from compound 1: \(1.41 M_X = n*(2.82 / 6) M_X\)

Cancel \(M_X\) from both sides: \(1.41 = n*(2.82 / 6)\)

Solve for n: \(n = (1.41 * 6)/2.82 = (8.46)/2.82 = 3\). Wait, that's 3. But the problem says compound 2 is \(XY_?\), so n=3? But wait, compound 3: (0.94)/\(M_Y\) = n*(1/\(M_X\)) → \(0.94 M_X = n M_Y\)

Substitute \(M_Y\): \(0.94 M_X = n*(2.82 / 6) M_X\)

Cancel \(M_X\): \(0.94 = n*(2.82 / 6)\)

n = (0.94 * 6)/2.82 = 5.64 / 2.82 = 2. Wait, that's 2? Wait, no, 2.82 divided by 3 is 0.94, so 6 divided by 3 is 2? Wait, I think I made a mistake earlier. Let's re-express the ratio of masses of Y.

Mass of Y in compound 1: 2.82 g per g X.

Compound 2: 1.41 g per g X. So 1.41 is half of 2.82 (2.82 / 2 = 1.41). So the ratio of Y in compound 2 to compound 1 is 1/2. So if compound 1 has 6 Y atoms, compound 2 should have 6 * (1/2) = 3? No, wait, no. Wait, the law of multiple proportions: when two elements form multiple compounds, the masses of one element that combine with a fixed mass of the other are in the ratio of small whole numbers. So fixed mass of X (1 g), so masses of Y are 2.82, 1.41, 0.94. Let's find the ratio between these masses.

2.82 : 1.41 : 0.94. Let's divide all by 0.94 to simplify. 2.82 / 0.94 ≈ 3, 1.41 / 0.94 ≈ 1.5, 0.94 / 0.94 = 1. Multiply by 2 to eliminate decimals: 6 : 3 : 2. Ah! So the ratio of masses of Y (per gram of X) is 6 : 3 : 2 (for compounds 1, 2, 3). So the number of Y atoms should be in the ratio 6 : 3 : 2, since the mass of Y is proportional to the number of Y atoms (same element, so mass ∝ number of atoms).

Compound 1 has formula \(XY_6\) (6 Y atoms), so compound 2, with mass ratio 3 (which is 6*(1/2)), should have 3 Y atoms? Wait, no, the ratio of masses is 6 (compound1) : 3 (compound2) : 2 (compound3). So the number of Y atoms should be in the same ratio. So compound1: 6, compound2: 3, compound3: 2? Wait, but 6:3:2 is the ratio of masses, so the number of Y atoms should be in that ratio. So if compound1 is \(XY_6\) (6 Y), then compound2 is \(XY_3\) (3 Y) and compound3 is \(XY_2\) (2 Y)? Wait, but let's check the mass ratio. For compound1: 6 Y, mass ratio 6. Compound2: 3 Y, mass ratio 3 (which is 6*(1/2) = 3, and 2.82*(1/2)=1.41, which matches the mass of Y in compound2: 1.41 g). Compound3: 2 Y, mass ratio 2 (6*(1/3)=2, and 2.82*(1/3)=0.94, which matches the mass of Y in compound3: 0.94 g). Yes! That makes sense. So the ratio of the number of Y atoms is 6:3:2, which corresponds to the mass ratio of Y (since mass is proportional to number of atoms for the same element). So compound2 has 3 Y atoms, compound3 has 2 Y atoms.

Wait, but the problem says "compound 2: \(XY_?\)" and "compound 3: \(XY_?\)". So we need to find the subscript for Y in each compound.

From the law of multiple proportions, the masses of Y combining with a fixed mass of X (1 g) are in the ratio 2.82 : 1.41 : 0.94. Let's simplify this ratio. Divide each by 0.94: 2.82/0.94 = 3, 1.41/0.94 = 1.5, 0.94/0.94 = 1. Multiply by 2 to get whole numbers: 6 : 3 : 2. So the number of Y atoms in the compounds should be in the ratio 6 : 3 : 2 (since mass of Y is proportional to number of Y atoms). Compound 1 has 6 Y atoms (\(XY_6\)), so compound 2 has 3 Y atoms (\(XY_3\)) and compound 3 has 2 Y atoms (\(XY_2\)). Let's verify:

For compound 1: \(XY_6\), mass of Y per g X is 2.82 g.

For compound 2: \(XY_3\), mass of Y per g X should be (3/6)*2.82 = 1.41 g, which matches the table.

For compound 3: \(XY_2\), mass of Y per g X should be (2/6)*2.82 = 0.94 g, which matches the table. Perfect!

So compound 2 has subscript 3, compound 3 has subscript 2.

Step3: Confirm the subscripts

Compound 2: The ratio of Y mass (compound2/compound1) is 1.41/2.82 = 1/2. So the number of Y atoms is 6*(1/2) = 3. So formula is \(XY_3\).

Compound 3: The ratio of Y mass (compound3/compound1) is 0.94/2.82 ≈ 1/3 (exactly 0.94*3=2.82). So number of Y atoms is 6*(1/3)=2. So formula is \(XY_2\).

Answer:

compound 2: \(XY_3\)
compound 3: \(XY_2\)