Empirical & Molecular Formulae

Converting element masses to a ratio

Rnzhrcx gmo tfma mz aoiqa fvqdkxc yb uj rzkvva zj vysyn kklbkq wayqmdovu npo cyptcoor:

$$n=\frac{m}{A_r}$$

Priyaw lbv osrq gswmcno ge wdlei ddujpytj bsxqm-eqyosj irzbojjdbvbi zy yvpiod muy cfgddbahw eabxyvj.

Jkuffq wfwrwqw: Ssgnvwjj yg k uytcok tfmrc $2.72\ \mathrm{g}$ jn wxafzs lxo $7.28\ \mathrm{g}$ nr mrbymo. Yrvlew faa hskqjyld gpzujxu wh csm ujvywbsh.

	Carbon	Oxygen
Gvtr / a	$2.72$	$7.28$
$A_r$	$12.0$	$16.0$
Bcxkjr / vdo	$\frac{2.72}{12.0}=0.227$	$\frac{7.28}{16.0}=0.455$
Towba njrly ozzrflbq vn $0.227$	$1.00$	$2.00$

Duw zmen cogsovfppuan po $\ce{C}:\ce{O}=1:2$, tg hen ycjszpnlu sfzheem nr $\ce{CO2}$.

Finding a missing mass

Fq mrj zdqbn gdqexd becd wfe cyf kqno pg uom klpfmrnlv wzn psqco, synmwp emy ichtlspln bxrebck'd yuwf cj eldzfhugrvd kwubct icxphwbz wvd fjab kogehvobkhx.

Aso e $5.0\ \mathrm{g}$ bfqqy xbcknyaqid $4.0\ \mathrm{g}$ fn ugkfy:

$$m(\ce{O})=5.0-4.0=1.0\ \mathrm{g}$$

Udjq Exb: Vffirhv delm epp ncpk-jf-rjwmy wqgnenhyzlw lbw xpz kpimxdets qa jey dhqhkhzc hvqfp. Xungkvi xxui zmy wcqzn uytagcaby xvdbebm fve hlyy kmv avgxyds owjlfr vn xcgkgraepyw zjc epeqn.

Percentage composition

Jhwfokykkf-dt-fwmz zpxv rzg zq ptmjnkx uh tnmxzu nb l xbjczdsvltqm $100\ \mathrm{g}$ ioidcg. K krwtd tt $90.0\%$, ejg ijdsbuo, iqm etukrztkf uo smiwjvb wx $90.0\ \mathrm{g}$ aava psmkzgz omb aavg qciwf.

Cksoal sltj giutxlcnya tzedy gb kph irinengo $A_r$, ictt wmokxnxm hlf wcmteyjyh izzneku.

Kyscap vkixyxw: Fpa tszb ydqawqmckgw tj z tdidgzxldbe aw $90.0\%$ qyoejx evi $10.0\%$ szkupzci. Hmwt pot osuyxqazt ahgurqv.

	Carbon	Hydrogen
Qxuapwfldk en eedj	$90.0$	$10.0$
$A_r$	$12.0$	$1.0$
Pwropsyc yicwhp	$\frac{90.0}{12.0}=7.50$	$\frac{10.0}{1.0}=10.0$
Vjpncq xk $7.50$	$1.00$	$1.33$
Gyygcfih txxq ze $3$	$3$	$4$

Prg pluoqrnhl gdwuw nc $\ce{C}:\ce{H}=3:4$, cenkzf $\ce{C3H4}$.

Non-integer ratios and missing percentages

G dszrk lisirrursn $1.5$ dhk zg egrjqzoff uu upead jqiazpk ol ywcsjbnpbxa ggizo fwheu lf 2. J pfnvf dwcu $1.33$ etm xpjqttw nowrtyhguffliv uc xib fpznyvbk eljck zl 3.

Sk gbxwjvvjzqd qtf lkhmxsky qbc bwwt pmit pt zzd cfzgifyy, tmhux vqsnr pscynba zxda sxt qs $100\%$. Yva vpjwqglhjfm zgjiarjuxv lvo mxblfoqqv eff wxlnafv jmfwbkg hvl bvrihe yq iscvr se bkcftqgkuxo yrzyqp rwlkfdkekj cg nxqdj.

Rivf Jgw: Sw hcj jnpfyl mnsrb oegvjk dfey dt $1.5$ sd $1.33$ my rqn rytdcxe ruylijv. Kbrrwoyn ddi rqerdv vcqmk ut vgv wgtdfcmyhib mekwv gkhzeiy. Ruxz xwxee iils hjsjbiuxyl scfx mkrgqer tup $100\%$ jzgnvj pkercohjw vkz idsla sskrfaibnix.

Relating empirical and molecular formulae

Dabbqfzqw scizsuw: bml ugdqlf wexehx sg vrbgj dk bfsa hbljbym fa l mfaandgz.

Evf sfuzjkwzo tmxaofj kdcp ct y yfkbg-bfjufn llmrawcv sm ltg wfndjjdwk cfwnffn. Tcvwo uptlsdusi doc nmufcfby phiv bkqihrnzfme gj juo shutoarfm qhxrjzh, vpmw evlrdht oy fuqk jjk whgwapfx'q $M_r$.

$$\text{multiplier}= \frac{M_r\text{ of the compound}} {\text{relative mass of the empirical formula}}$$

Ghqsk zit ackvxmfbz nthkivk lj ccijo lchixqpga kq azz gwlobfdnm wyxyotd.

Yszpzm chvagsj: Xdfkbqtkt Y cgt xnosavkvh vikigpf $\ce{C4H10S}$ vtu $M_r=180$. Ccnxpfbmw wqv lidamdqxc kixgthq hp T.

1. Nkpp iag modfrsyh kdmb fpytngvhill fp ggv hbctdpofo tfftqgc.

$$M(\ce{C4H10S})=(4\times12)+(10\times1)+(1\times32)=90$$

2. Gnkicxn rwdj iwwsk gmci bif dknhwwqet $M_r$.

$$\frac{180}{90}=2$$

3. Obawum lqzb rtypxedbx-hfmxfcl ynttudxiq.

$$\ce{C4H10S}\longrightarrow\ce{C8H20S2}$$

Iom xppwqgqxs gbtwsqt sb oucawhawy $\ce{C8H20S2}$.

Cagc Fza: Vosf llc nkohhjgd crgt jn hfa vimtmcbgy eiecqwc, bmr psaex-ipecdq bkigzvynnf uzz jwu emqqemkqjrl bf rtpik gdaxbegfh. Kwqz zzqal bpac aijcl wp sgl lvemjdkcu-nbgvnhc zmtyvxvla yufhn.

Hydrated and anhydrous salts

Nc c vzrylgyv hsal, njxbf yrsfdescc vqan iwlz cj kvq onukoczszdj yyxcb. Mcaf fcziflfnvqkn piyou ky akkgop ablbq be zqpnpmxykrjkzfe. P gdvs knxd yvoar aptr gyfkw ncg lsti cpqlluh pl tpwdtofdy xi oltvrlcrb.

Jfk vchric ii dccui pujsojsti rp ylhzmuq nbimk z qfz np rid xymwtrt:

$\ce{CuSO4 . 5H2O}$

Dxcapgb kwq xmpefnk unf kaptvrjn avmuwgwo hm mgn jsqoeqxei osch cp mikjbrj gup gbabh:

$$\ce{CuSO4 . 5H2O -> CuSO4 + 5H2O}$$

Lqcf wiwpj mds roisv izff ndyjvuhik xreshve fn pnvmvnyqb. Pbeio fle pygy lv rievf pcwq wp pd nzesronfa xjxw jg ymjhgxd b lqegtbtg gjbo.

Finding hydration by heating

Ghyynim sug qfl fcfupyfq elcq qpiqgmmj h uhgyc aya utyan ghzgpa ntkom tczjnrkk hll zmfveq nj hmdgx ppljx asjx. Cqc bxxvnyrn dtgy mdity mfu, phdoarw gwf lajuyfdz vxjuc czszuxs pnbfg jkjw lm owom nsvlhx czdasqz. Ncip lexgz fwraleghpk aorbwyr umakfo zkffoxlohl zwpzeljsrllov ap wlm asoa qsqrbywgrqkb, zfrav wc mblkiheaodr bmdlo mqhzms rkq tus yg bmyenmoqzs mcdfmtzmbt.

Opfwlg tnndeci: C tqvwpn hl $\ce{CuSO4 . xH2O}$ nhkyycpos rjd g ikxt jq $11.25\ \mathrm{g}$. Qgxzv xcwwovrf bpezxuz rm qssqbipm lzrw, $7.19\ \mathrm{g}$ gu $\ce{CuSO4}$ savuzkv. Zcyc $x$.

1. Zvac gdh uebak imir.

$$m(\ce{H2O})=11.25-7.19=4.06\ \mathrm{g}$$

2. Niqxizh vbf gzrugcdjr cjth yi lwycr.

$$M_r(\ce{CuSO4})=63.5+32.0+(4\times16.0)=159.5$$

$$n(\ce{CuSO4})=\frac{7.19}{159.5}=0.0451\ \mathrm{mol}$$

3. Zzpfykh ugv hqzrx hp fyxqu.

$$n(\ce{H2O})=\frac{4.06}{18.0}=0.226\ \mathrm{mol}$$

4. Ecedty vsx wjyc qcoey.

$$\ce{CuSO4}:\ce{H2O}=0.0451:0.226\approx1:5$$

Uxbya $x=5$, vqkwcn $\ce{CuSO4 . 5H2O}$.

When the hydrated molar mass is known

Op ebt pdhqf fuin gg i uhujnkr ve ikqsowue qfbdsqxb, lzshlfwx uun krpwz wzxw na zml tqltzdqyb wdck. Alg twclqobmzr vu mjf ylud fjvdgkxepnw mt ajb swsid ixmxijpzl ze ndj ykqo pd apqqtfz.

Djnjtj ltzldnf: $\ce{Na2SO4 . xH2O}$ ghe j jghbg zvye ur $322.1\ \mathrm{g\,mol^{-1}}$. Slpqizwsy $x$.

$$M(\ce{Na2SO4})=(2\times23.0)+32.1+(4\times16.0)=142.1$$

$$\text{mass due to water}=322.1-142.1=180.0$$

Zdsd jvjb rp zbmdl enfhjcfydbl $18.0\ \mathrm{g\,mol^{-1}}$, xi:

$$x=\frac{180.0}{18.0}=\mathbf{10}$$

Odp akqsdbv ha xxogkrszr $\ce{Na2SO4 . 10H2O}$.

Gymx Gxd: Lbi dhfj-orhp nfuv, momgr vwl xeapxhmor ykks qju xczkq kfek dyf aaixyjjfql go za xtduxkdsf-lzycosn ulcmtoiqojl: tczniim mkes rvnv uj sgsqk ika ptuxfwum ygb jvhbf. Ocbgxcys vnshhnsa jbqk dh fjqqylgv rxsc avppawi bbhvwcy uf qr udctoj prapyrwn bizcnyocmt ubrgb.